Exploration 1: Commutativity

November 7, 2023.

Questions:

What characterizes commutative elements of a group?
What characterizes commutative subsets of a group?
How do we make a group abelian?
How do we make a group not abelian?
What are quotient groups?

When taking products of group elements, like $gh$ , the ordering matters. In other words, $gh=hg$ is not true in general, unless $g$ or $h$ is the identity element $e$ . We saw this with the symmetric group, where the ordering of composing permutations matters.

In this exploration, we’ll look at all the ways $gh=hg$ can actually be true. That is, we’ll explore how we can actually make $g$ commute with $h$ .

In this section, we look at elements that commute with everything.

Every group has an identity element $e$ , and the identity element always commutes with any element: $eg=g=ge$ . In the worst case, the identity element is the only element that commutes with every element.

On the flipside, abelian groups are exactly those groups where every element commutes with every element. So at maximum, every element commutes with every element. The cyclic group is a good example of this.

When some given element $g$ commutes with every element in the group, we say that $g$ is central. The central elements of a group $G$ are collectively referred to as the center of $G$ , written $Z(G)$ . As discussed, the center is always in between the minimum “just the identity element” (a trivial center) and the maximum “every element” (in the case of abelian groups.)

Example: The group of integers under addition is abelian.

Integer addition is commutative, so $g+h=h+g$ for all $g,h\in\ZZ$ .

Example: The symmetric group with

n\ge 3

elements has a trivial center.

To be in the center, a permutation $\sigma$ must commute with every permutation.
In particular, $\sigma$ must commute with $(a~b)$ , so we have $\sigma(a~b)=(a~b)\sigma$ , which implies $\sigma$ either fixes $a,b$ or swaps $a,b$ .
Using the same logic for $(b~c)$ , $\sigma$ either fixes $b,c$ or swaps $b,c$ . But $\sigma$ can’t swap $b,c$ since that would mean $\sigma$ neither fixes nor swaps $a,b$ (using the assumption that $n\ge 3$ so that $a,b,c$ are distinct.) So the only option is that $\sigma$ fixes $a,b,c$ .
Applying this argument inductively to all elements $a,b,c,d,\ldots$ we show that $\sigma$ must fix every element in order to commute with all these $2$ -cycles. Therefore, $\sigma$ can only be the identity permutation.

Example: The center of the group of invertible

n\times n

matrices under matrix multiplication are exactly scalar multiples of the identity matrix.

Let $Z$ be a central matrix, so that it commutes with every invertible matrix in the group.
In particular, it commutes with the matrix $E=I+E_{ij}$ (where $i\ne j$ ), which is the identity matrix except with a $1$ at an off-diagonal entry at row $i$ column $j$ . For example: $E=\left[\begin{matrix}1&0&0\\0&1&1\\0&0&1\end{matrix}\right]\quad(\text{for }n=3,i=2,j=3)$ This matrix $E$ is a member of our group because it is invertible (it is triangular and therefore $\det(E)$ is the product of the diagonal, which is $1$ , thus invertible.)
Note that by matrix multiplication, $(ZE)_{ik}$ is equal to the inner product of (the $i$ th row of $Z$ ) and (the $k$ th column of $E$ ), which by construction is exactly $(ZE)_{ik}=\begin{cases}Z_{ik}+{\color{red}0}&\text{ for }k\ne j\\Z_{ik}+{\color{red}Z_{ii}}&\text{ for }k=j\end{cases}$ $ZE=Z(I+E_{ij})=Z+\left[\begin{matrix}Z_{1,1}&Z_{1,2}&Z_{1,3}\\Z_{2,1}&Z_{2,2}&Z_{2,3}\\Z_{3,1}&Z_{3,2}&Z_{3,3}\end{matrix}\right]\left[\begin{matrix}0&0&0\\0&0&1\\0&0&0\end{matrix}\right]=Z+\left[\begin{matrix}0&0&Z_{1,2}\\{\color{red}0}&{\color{red}0}&{\color{red}Z_{2,2}}\\0&0&Z_{3,2}\end{matrix}\right]$ Similarly, $(EZ)_{kj}=\begin{cases}Z_{kj}+{\color{blue}0}&\text{ for }k\ne i\\Z_{kj}+{\color{blue}Z_{jj}}&\text{ for }k=i\end{cases}$ $EZ=(I+E_{ij})Z=Z+\left[\begin{matrix}0&0&0\\0&0&1\\0&0&0\end{matrix}\right]\left[\begin{matrix}Z_{1,1}&Z_{1,2}&Z_{1,3}\\Z_{2,1}&Z_{2,2}&Z_{2,3}\\Z_{3,1}&Z_{3,2}&Z_{3,3}\end{matrix}\right]=Z+\left[\begin{matrix}0&0&{\color{blue}0}\\Z_{3,1}&Z_{3,2}&{\color{blue}Z_{3,3}}\\0&0&{\color{blue}0}\end{matrix}\right]$
Since $ZE=EZ$ due to $Z$ being in the center, equate the above at index $i,j$ to obtain $\begin{aligned} (ZE)_{ij}&=(EZ)_{ij}\\ Z_{ij}+{\color{red}Z_{ii}}&=Z_{ij}+{\color{blue}Z_{jj}}\\ {\color{red}Z_{ii}}&={\color{blue}Z_{jj}}\\ \end{aligned}$ for all $i\ne j$ . This implies all the diagonal entries of $Z$ are equal.
As for the off-diagonal entries, note that $ZE_{ij}$ is $Z_{ki}$ only on column $j$ and zero otherwise, and $E_{ij}Z$ is $Z_{jk}$ only on row $i$ and zero otherwise. That means for $k\ne i$ , $(ZE)_{kj}=(EZ)_{kj}\implies Z_{kj}+Z_{ki}=Z_{kj}+0\implies Z_{ki}=0$ and for $k\ne j$ , $(ZE)_{ik}=(EZ)_{ik}\implies Z_{ik}+0=Z_{ik}+Z_{jk}\implies Z_{jk}=0$ meaning that any off-diagonal entry of $Z$ must be zero, i.e. $Z$ is diagonal.
Since $Z$ is diagonal with every diagonal entry equal, $Z$ must be a scalar multiple of the identity matrix $\lambda I$ . So every central matrix must be in the form $\lambda I$ .
Conversely, all matrices in the form $\lambda I$ are central because $(\lambda I)M=\lambda MI=M(\lambda I)$ for an arbitrary matrix $M$ .
Therefore the center of the group of invertible matrices consists of exactly the scalar multiples of the identity matrix.

In this section, we learn about subsets that commute with everything.

Being a central element is a rather strong property. A central element must commute with every element.

Instead of considering single elements that commute with all elements in a group, consider subsets that commute with all elements in a group. Let $S\subseteq G$ be a subset of $G$ , and let $Sg$ (resp. $gS$ ) represent the result of right-multiplying (resp. left-multiplying) all of $S$ by some $g\in G$ . Then if $Sg=gS$ for all $g\in G$ , we can characterize $S$ as something like being a “central” subset of $G$ . How do its properties differ from those of central elements?

Notice that since $Sg=gS$ implies $S=gSg^{-1}$ for all $g\in G$ , $S$ has the property that for any of its elements $s\in S$ , then the element $gsg^{-1}$ is also in $S$ , for every $g\in G$ .

This map $s\mapsto gsg^{-1}$ is called conjugation by $g$ . When $t=gsg^{-1}$ , we say that $t$ is a conjugate of $s$ by $g$ .

So our earlier property that $Sg=gS$ for all $g\in G$ can be rewritten as $S=gSg^{-1}$ for all $g\in G$ . Since the idea is that $S$ is left unchanged after conjugation by arbitrary $g$ , we call this property invariance under conjugation. Specifically, a set $S$ or element $x$ is invariant under conjugation when conjugation by every $g\in G$ leaves $S$ or $x$ unchanged, i.e. $gSg^{-1}=S$ or $gxg^{-1}=x$ .

Theorem: The central elements are exactly the elements invariant under conjugation.

$z$ is central iff $zg=gz$ for all $g\in G$ .
$z$ is invariant under conjugation iff $z=gzg^{-1}$ for all $g\in G$ .
But again, $zg=gz$ and $z=gzg^{-1}$ imply each other, so these conditions are equivalent.

Theorem: The subsets

S

that commute with every

g\in G

are exactly those that are invariant under conjugation.

To commute with every $g\in G$ , $S$ must satisfy $Sg=gS$ for all $g\in G$ .
To be invariant under conjugation, $S$ must satisfy $S=gSg^{-1}$ for all $g\in G$ .
But $Sg=gS$ and $S=gSg^{-1}$ imply each other, so these conditions are equivalent.

Corollary: The union of two subsets

S,T

invariant under conjugation is also invariant under conjugation.

Since group product distributes over set union, we have $(S\cup T)g=Sg\cup Tg$ for all $g\in G$ . But since $S,T$ are invariant under conjugation, we have from the previous theorem that both $S,T$ commute with every element $g\in G$ , and therefore $(S\cup T)g=Sg\cup Tg=gS\cup gT=g(S\cup T)$ implying that $S\cup T$ commutes with everything and is therefore invariant under conjugation.

So if we want to study subsets that are “central”, we can think about constructing subsets that are invariant under conjugation. The obvious way to create such subsets is to take every conjugate of some element $g\in G$ to arrive at the subset $[g]$ . Since every conjugate of $g$ is in $[g]$ , $[g]$ must be invariant under conjugation. We can prove that more rigorously:

Theorem: The set

[g]

of all conjugates of

g\in G

is invariant under conjugation.

Since every element in $C$ is a conjugate of $g$ by some element $h\in G$ , we can represent every element of $[g]$ as $hgh^{-1}$ .
To show that conjugating this element (by arbitrary $k\in G$ ) gives the element in $[g]$ , note that $k(hgh^{-1})k^{-1}=(kh)g(kh)^{-1}\in [g]$ , i.e conjugating an arbitrary element $hgh^{-1}\in [g]$ by an arbitrary element $k\in G$ results in an element $(kh)g(kh)^{-1}$ in $[g]$ , thus $[g]$ is invariant under conjugation.

Thus every element $g\in G$ gives rise to a subset $[g]\subseteq G$ that contains exactly all the conjugates of $g$ , which automatically makes it invariant under conjugation.

In this section, we discuss equivalence relations.

Imagine doing this process for every element in $G$ , so that for every element $g\in G$ you can identify a subset $[g]$ that $g$ belongs to.

In fact, every element $g$ belongs to exactly one subset $[g]$ . To see this, note that if $g\in [g]$ was also in a second subset $[h]$ , then $g$ must be a conjugate of $h$ . But since $[g],[h]$ must both be invariant under conjugation (as we just proved), every conjugate of $g$ must also be in $[h]$ (so $[g]\subseteq [h]$ ), and every conjugate of $h$ must also be in $[g]$ (so $[h]\subseteq [g]$ .) This implies $[g]=[h]$ .

Then the subsets $[g]$ collectively partition the set $G$ . A partition of a set is defined as any grouping of all elements into subsets such that each element belongs to exactly one subset.

The reason that conjugation gives rise to a partition is due to three essential properties:

Conjugation is transitive: if $a,b$ are conjugate by $h$ , and $b,c$ are conjugate by $k$ , then $a,c$ are conjugate by $kh$ . (We actually proved this earlier.) This ensures that if $g\in [h]$ , then $g$ is conjugate to $h$ and everything $h$ is conjugate to, thus $[h]\subseteq [g]$ .
Conjugation is symmetric: if $g$ is conjugate to $h$ , then $h$ is conjugate to $g$ . This ensures that in the previous scenario, $h$ is also conjugate to everything $g$ is conjugate to, thus $[g]\subseteq [h]$ . Together with the previous result, this implies $[g]=[h]$ . Thus $g\in [h]$ implies $[g]=[h]$ , meaning $g$ cannot be in more than one distinct subset.
Conjugation is reflexive: every element is conjugate to itself by $e$ . This ensures that every element belongs to some subset, i.e. every element $g$ belongs to at least one subset. Together with the previous result, this means every element belongs to exactly one subset.

So any relation that is transitive, symmetric, and reflexive gives rise to a partition of the underlying set into these subsets $[g]$ . Such a relation is called an equivalence relation, often denoted $\sim$ . The partitions $[g]$ are called equivalence classes, each containing all elements equivalent to its representative $g$ by the given equivalence relation $\sim$ .

Like all equivalence relations, conjugation partitions the group into equivalence classes, called conjugacy classes. Each conjugacy class $[g]$ contains all elements $hgh^{-1}$ conjugate to $g$ .

In this section, we learn how to determine the size of conjugacy classes.

The size of a conjugacy class $|[g]|$ is the number of distinct elements in the form $hgh^{-1}$ for some $h\in G$ . How do we determine the number of distinct conjugates of $g$ ?

To think about this problem, imagine taking the conjugate of $g$ by every element in $G$ , so you have potentially $|G|$ conjugates of $g$ . Whenever two conjugates of $g$ are equal, $aga^{-1}=bgb^{-1}$ , then they are two ways to write the same conjugate. Note that this condition $aga^{-1}=bgb^{-1}$ is trivially an equivalence relation $a\sim b$ , since it’s based on equality:

Reflexivity: $aga^{-1}=aga^{-1}$ .
Symmetry: $aga^{-1}=bgb^{-1}$ implies $bgb^{-1}=aga^{-1}$
Transitivity: if $aga^{-1}=bgb^{-1}$ and $bgb^{-1}=cgc^{-1}$ , then $aga^{-1}=cgc^{-1}$ .

NB: This $\sim$ is a different equivalence relation than the one we gave for conjugacy.

Being an equivalence relation, $\sim$ divides $G$ into equivalence classes, where two elements $a,b$ are in the same equivalence class iff conjugating $g$ by $a$ and by $b$ results in the same element. Since each equivalence class represents a distinct conjugate of $g$ , the number of distinct conjugates of $g$ is equal to the number of equivalence classes.

So, how do we find the number of equivalence classes?

Here is the key insight: the equality $aga^{-1}=bgb^{-1}$ can be rewritten as $(b^{-1}a)g=g(b^{-1}a)$ meaning two conjugates of $g$ are equal every time the element $b^{-1}a$ commutes with $g$ . Let $C_G(g)$ denote the set of such elements in $G$ that commute with $g$ , called the centralizer of $g$ . Then the condition $aga^{-1}=bgb^{-1}$ simplifies to the condition $b^{-1}a\in C_G(g)$ , which further simplifies to $a\in bC_G(g)$ . This directly characterizes the equivalence classes: every equivalence class $[a]$ under $\sim$ is in the form $bC_G(g)$ for some $b\in G$ .

There is a bijection between any two equivalence classes $bC_G(g)$ and $aC_G(g)$ . Simply left-multiply $bC_G(g)$ by $ab^{-1}$ to obtain $aC_G(g)$ . This is a bijection because there exists an inverse: left-multiplying by $ba^{-1}$ . Why is this important? Because the existence of a bijection between every equivalence class implies that every equivalence class has the same size: $|C_G(g)|$ .

If the size of every equivalence class is $|C_G(g)|$ , then the number of equivalence classes is $|G|/|C_G(g)|$ . Since each equivalence class represents a distinct conjugate of $g$ , we have obtained the number of distinct conjugates of $g$ : $|[g]|=|G|/|C_G(g)|$

Thus the size of the conjugacy class $[g]$ is the order of the group divided by the size of $g$ ’s centralizer. This reduces the problem to finding the size of a centralizer. For example:

Example: The centralizer of every element in an abelian group is the group itself.

Since everything commutes with everything in an abelian group, the centralizer of every element consists of the whole group.

Corollary: Abelian groups are exactly the groups where every conjugacy class is of size $|G|/|C_G(g)|=1$ .

We can also do this for permutations and matrices, but please skip these if the domain is not familiar to you!

Example: The centralizer of a permutation

\sigma

S_n

consists of all permutations that preserve the cycle structure of

\sigma

, of which there are

\prod_j j^{N_j}N_j!

(where

N_j

denotes the number of disjoint cycles of length

j

\sigma

Express $\sigma$ in disjoint cycle notation so that we can take an arbitrary cycle $(a_1~a_2~\ldots~a_n)$ .

If $\tau$ is in the centralizer of $\sigma$ , then it commutes with $\sigma$ . In particular, for each $a_i$ , we have $\tau(\sigma(a_i))=\sigma(\tau(a_i))$ , which simplifies to $\tau(a_{i+1})=\sigma(\tau(a_i))$ . Because this equation implies that $\sigma$ takes $\tau(a_i)$ to $\tau(a_{i+1})$ , we know that $\tau(a_i)$ and $\tau(a_{i+1})$ must be in the same cycle. Using the same argument with $\tau^{-1}$ (which also commutes with $\sigma$ ), we know $\tau^{-1}(a_i)$ and $\tau^{-1}(a_{i+1})$ must be in the same cycle.

If $\tau$ takes elements in the same cycle to the same cycle, and the preimage $\tau^{-1}$ also takes elements of the same cycle to the same cycle, then any $\tau$ in the centralizer of $\sigma$ simply permutes the elements within each cycle of $\sigma$ . This means each cycle in $\sigma$ is mapped to a cycle of the same length, thus preserving the cycle structure of $\sigma$ .

The number of such $\tau$ is the number of such permutations, which is the product of the number of permutations of each cycle of $\sigma$ . If there are $N_j$ cycles of length $j$ in $\sigma$ , then the number of $\tau$ (the size of the centralizer of $\sigma$ ) is equal to $|C_{S_n}(\sigma)|=\prod_j j^{N_j}N_j!$

Corollary: The size of the conjugacy class of a permutation $\sigma$ in $S_n$ is $|G|/|C_G(\sigma)|=n!/\prod_j j^{N_j}N_j!$ (where $N_j$ denotes the number of disjoint cycles of length $j$ in $\sigma$ .)

Example: The centralizer of a diagonalizable matrix

M

in the group of invertible

n\times n

matrices under matrix multiplication consists of all matrices block-diagonalizable in the same eigenbasis of

M

, of which there are

\prod_j |F|^{n_j^2-n_j}

(where

n_j

denotes the multiplicity of the

j

th distinct eigenvalue, and

|F|

denotes the size of the field that the matrices are defined over.)

A diagonalizable matrix $M$ is one that can be represented as $M=PDP^{-1}$ where $D$ is a diagonal matrix and the columns of $P$ form an eigenbasis of $M$ . To find the centralizer, we’re trying to find all matrices $A$ that commute with $PDP^{-1}$ , i.e. $APDP^{-1}=PDP^{-1}A$
Let $A=PBP^{-1}$ for some $B$ . Then we have $PBP^{-1}PDP^{-1}=PDP^{-1}PBP^{-1}$ which simplifies to $BD=DB$
Thus $B$ must commute with the diagonal matrix $D$ . By definition of matrix multiplication, $(BD)_{ij}$ must be equal to $\sum_k B_{ik}D_{kj}=B_{ij}D_{jj}$ , since $D$ is diagonal meaning $D_{kj}$ is zero everywhere except when $k=j$ . Likewise, $(DB)_{ij}=\sum_k D_{ik}B_{kj}=D_{ii}B_{ij}$ . Since $BD=DB$ we get $(BD)_{ij}=(DB)_{ij}\implies B_{ij}D_{jj}=D_{ii}B_{ij}\implies B_{ij}D_{jj}=B_{ij}D_{ii}$ .
$B_{ij}D_{jj}=B_{ij}D_{ii}$ is trivially true if $i=j$ . For $i\ne j$ , it means that $B_{ij}$ is nonzero only if $D_{ii}=D_{jj}$ . This means $B$ has nonzero entries only within blocks where the eigenvalues $D_{ii}$ are equal. In other words, $B$ is block-diagonal, with each block corresponding to the eigenspace of each distinct eigenvalue.
$B$ is block-diagonal, so $A=PBP^{-1}$ is block-diagonalizable in the same eigenbasis $P$ of $M$ . Therefore, the matrices $A$ that commute with $M$ are exactly the ones block-diagonalizable in the same eigenbasis of $M$ .
Thus the size of the centralizer of $M$ can be computed as the product of the number of possible matrices for each block in the block diagonal form of $B$ .
Then we can count then number of matrices by counting the number of possible $B$ . Each $n\times n$ block in $B$ has $n^2-n$ entries (all entries but the diagonal) that can be filled with any value, therefore each $n\times n$ block contributes $|F|^{n^2-n}$ possibilities where $|F|$ denotes the size of the field that the matrices are defined over. Blocks are determined by the multiplicity of eigenvalues of $M$ , thus we take the product $\prod_j |F|^{n_j^2-n_j}$ where $n_j$ denotes the multiplicity of the $j$ th distinct eigenvalue.

Corollary: The size of the conjugacy class of a diagonalizable matrix $M$ in the group of invertible $n\times n$ matrices under matrix multiplication is $|G|/|C_G(M)|=(\prod_{i=0}^{n-1} |F|^n-|F|^i)/(\prod_j |F|^{n_j^2-n_j})$ (where $n_j$ denotes the multiplicity of the $j$ th distinct eigenvalue, and $|F|$ denotes the size of the field that the matrices are defined over.)

In this section, we learn about the relationship between conjugacy classes and central elements.

Here’s an easy-to-prove fact:

Theorem: Central elements are exactly the elements that are invariant under conjugation.

Being central means $gz=zg$ for every $g\in G$ , and being invariant under conjugation means $z=gzg^{-1}$ for every $g\in G$ . But $gz=zg\iff z=gzg^{-1}$ .

Corollary: The conjugacy class

[z]

of a central element

z

is always a singleton set, and the representative

z

of a singleton conjugacy class

[z]

is a central element.

The conjugacy class of $z$ contains all elements conjugate to $z$ . But since $z$ is invariant under conjugation, as we just proved, the only element conjugate to $z$ is $z$ itself. Therefore $z$ is in a singleton conjugacy class.

Conversely, if $z$ is in a singleton conjugacy class, it means only $z$ is conjugate to $z$ , therefore $z$ is invariant under conjugation, therefore central.

In particular, the identity element $e$ (which is always central) is always in a singleton conjugacy class.

An important result that arises from this is that we can split a group’s conjugacy classes into two types: the central conjugacy classes (which are the singleton conjugacy classes) and the non-central conjugacy classes. This relationship is described below: $|G|=|Z(G)|+\sum_i|[g_i]|$ where $|[g_i]|$ denotes the size of the $i$ th non-central conjugacy class. Using what we learned earlier, we can rewrite this as $|G|=|Z(G)|+\sum_i|G|/|C_G(g_i)|$

The result above is known as the class equation of a group, and is used in many number-theoretic proofs about the center. For example:

Theorem: Every group with prime power order (a $p$ -group) has a non-trivial center, whose order is divisible by

p

$p$ -groups are of prime power order, so $|G|=p^n$ where $n\ge 1$ .
Using the fact that $|[g_i]|=|G|/|C_G(g_i)|$ , we know that the size of the conjugacy class of $g_i$ must be a factor of $|G|$ , i.e. it must be a prime power $p^{k_i}$ where $k_i\le n$ .
Then we can write the class equation as $p^n=|Z(G)|+\sum_i p^{k_i}$ Since the LHS is divisible by $p$ , so must the RHS. The sum $\sum_i p^{k_i}$ is divisible by $p$ , because each of its terms is divisible by $p$ . Then in order for the RHS as a whole to be divisible by $p$ , $|Z(G)|$ must also be divisible by $p$ .
But if $|Z(G)|$ is divisible by $p$ then it is not $1$ , therefore the center is non-trivial.

In this section, we learn about subgroups.

A subgroup of a group $G$ is a subset of $G$ that is also a group. In other words, it is a subset of $G$ that includes identity and is closed under product and inverse.

Every element $g$ generates a subgroup $\<g\>$ by taking powers of itself: $\<g\>=\{\ldots,g^{-3},g^{-2},g^{-1},e,g,g^2,g^3,\ldots\}$ . We already know this as the cyclic group generated by $g$ . You can also generate subgroups from multiple elements by taking all products and inverses: $\<g,h\>=\{e,g,h,g^2,gh,hg,h^2,g^3,\ldots\}$

In fact, some of the subsets we’ve been working with are actually subgroups, as proven below:

Theorem: The center

Z(G)

is a subgroup of

G

The identity $e$ is always central, so $e\in Z(G)$ .
The product of two central elements $g_1,g_2\in Z(G)$ $g_{1}, g_{2} \in Z (G)$ is central.
- Proof: $(g_1g_2)h=g_1hg_2=h(g_1g_2)$ for all $h\in G$ , thus $g_1g_2\in Z(G)$ .
The inverse of a central element $g\in Z(G)$ $g \in Z (G)$ is central.
- Proof: $\begin{aligned}gh&=hg\\g^{-1}ghg^{-1}&=g^{-1}hgg^{-1}\\hg^{-1}&=g^{-1}h\end{aligned}$ for all $h\in C_G(g)$ , thus $h^{-1}\in C_G(g)$ .
Thus $Z(G)$ forms a group, and is a subgroup of $G$ .

Theorem: The centralizer

C_G(S)

is a subgroup of

G

Recall that to be in the centralizer, every $g\in C_G(S)$ must commute with $S$ .
$C_G(S)$ has identity: Clearly $e\in C_G(S)$ because $eS=S=Se$ .
$C_G(S)$ has inverses: If $g\in C_G(S)$ then for all $h\in G$ we have $hg=gh$ , which is the same as saying $gh^{-1}=h^{-1}g$ , therefore $g^{-1}\in C_G(S)$ .
$C_G(S)$ has products: If $g,h\in C_G(S)$ , then for all $k\in G$ we have $gk=kg$ and $hk=kg$ . Then $(gh)k=gkh=k(gh)$ , thus $gh\in C_G(S)$ .
Since it has identity, inverses, and products, the centralizer $C_G(S)$ is always a subgroup of $G$ .

Theorem: The intersection of two subgroups is a subgroup of both.

Both subgroups contain $e$ , so their intersection contains $e$ .
Both subgroups are closed under product, so their intersection is closed under product.
- To make this more clear, this is because the product of two elements that are contained in both subgroups must remain contained in both subgroups by definition of subgroup.
Both subgroups are closed under inverses, so their intersection is closed under inverses.
Since the intersection is a subset of both given subgroups, contains identity, and is closed under product and inverses, it is a subgroup of both.

Since the center is the intersection of all centralizers of a group, the above theorem provides an alternate proof that the center is a subgroup (using the fact that all centralizers are subgroups.)

In this section, we make a group the least abelian it can be.

To make a group the least abelian it can be, we want it to have the smallest possible center. A group whose center is trivial is called centerless.

One way to make the center of a group smaller is to map every central element $z\in Z(G)$ to the identity element $e$ . Let $\pi$ be this map.

The problem is, if $az=b$ , then when we map $z$ to $e$ we get $a=b$ under $\pi$ . So when defining $\pi$ , not only do we need to map central elements to $e$ , we need to collapse elements differing by some $z$ into the same element. How do we achieve this definition? Equivalence relations!

Say $a\sim b$ when $a,b$ differ by a central element, so that $a,b$ are in the same equivalence class exactly when they differ by a central element (and therefore $a=b$ under $\pi$ .) Then as usual, $\sim$ partitions $G$ into equivalence classes $[g]$ , and as each class represents a distinct element under $\pi$ , we can define $\pi$ as sending every element $g\in G$ to its corresponding equivalence class $\pi(g)$ . This accomplishes the goal of mapping central elements to an identity element (the equivalence class of $e$ ), while collapsing elements differing by some central elements to the same equivalence class.

Well, that’s the plan, anyways. We don’t yet have any idea what the structure of $\pi(g)$ is, much less a guarantee that it’s a group.

To ensure it’s a group, note that for $a,b$ to differ by a central element, we can say $az=b$ for some $z\in Z(G)$ . Let’s shorten that condition to $b\in aZ(G)$ . To explain: $aZ(G)$ contains every element that differs from $a$ by some central element, and $b$ is one of them iff $a,b$ differ by some central element.

But the equivalence relation $a\sim b\iff b\in aZ(G)$ just checks for set membership. In other words, we have essentially $b\in [a]\iff b\in aZ(G)$ , meaning that we can identify $aZ(G)$ itself as the equivalence class $[a]$ ! Thus the following manipulations are valid: $g[a]=g(aZ(G))=(ga)Z(G)=[ga]$ $[a][b]=aZ(G)bZ(G)=(ab)Z(G)=[ab]$ $[g]^{-1}=Z(G)^{-1}g^{-1}=g^{-1}Z(G)=[g^{-1}]$ Note that this is only possible since $Z(G)$ commutes with every element by definition of being the center, and that $Z(G)$ is a subgroup of $G$ : $Z(G)Z(G)=Z(G)$ because subgroups are closed under product and $Z(G)^{-1}=Z(G)$ because subgroups are closed under inverse.

Armed with these manipulations and the knowledge that $\pi(a)=[a]=aZ(G)$ , let’s ensure that the image of $\pi$ is indeed a group.

Identity: $\pi(e)=[e]$ satisfies the identity laws $[g][e]=[ge]=[g]=[eg]=[e][g]$ .
Closed under product: $\pi(g)\pi(h)=[g][h]=[gh]=\pi(gh)$ .
Closed under inverses: $\pi(g)^{-1}=[g]^{-1}=[g^{-1}]=\pi(g^{-1})$ .

So $\im\pi$ is indeed a group!

Above, we used a number of mechanisms to arrive at a new group by sending the center to $e$ . Here’s a summary:

First, we defined an equivalence relation predicated on differing by a central element.
We used this equivalence relation to partition the group into equivalence classes $[g]$ .
By expressing the equivalence relation as set membership, we found that each of the equivalence classes $[g]$ is exactly the set $gZ(G)$ .
Using properties of the center, we proved that the set of all $gZ(G)$ for all $g\in G$ form a group.

This overall operation, of sending a subgroup to $e$ to obtain a new group, is called quotienting the group $G$ by the subgroup $Z(G)$ . The resulting group of equivalence classes is known as a quotient group, and in this case, it is denoted $G/Z(G)$ . (reads as “ $G$ mod the center of $G$ .”)

This quotient group gives rise to one of the most efficient ways to tell if a group is abelian.

Theorem:

G

is abelian iff

G/Z(G)

is cyclic.

If $G$ is abelian, then $G=Z(G)$ by definition, so $G/Z(G)$ sends every element of $G=Z(G)$ to the identity $e$ . Thus $G/Z(G)$ is trivial, therefore cyclic. To show the other direction, assume $G/Z(G)$ is cyclic, generated by some generating equivalence class $[g]=gZ(G)$ .

Every element of the cyclic group $G/Z(G)$ is expressible as a power of the generator $(gZ(G))^k=g^kZ(G)$ . Each of these equivalence classes $g^kZ(G)$ consists of products of $g^k$ with each element in $Z(G)$ . Since the equivalence classes cover the entire group $G$ , every element of $G$ is expressible as some element $g^kz$ where $z\in Z(G)$ .

But then two arbitrary elements $g^{k_1}z_1,g^{k_2}z_2$ must commute via $(g^{k_1}z_1)(g^{k_2}z_2)=z_2g^{k_1}g^{k_2}z_1=z_2g^{k_1+k_2}z_1=z_2g^{k_2}g^{k_1}z_1=(g^{k_2}z_2)(g^{k_1}z_1)$ thus $G$ is abelian.

In this section, we formalize quotient groups.

In general, given a subgroup $H\le G$ , we can attempt to find the quotient group $G/H$ by doing the following:

Since we want to send all elements of $H$ to $e$ , we want to consider two elements equivalent if they differ by any element $h\in H$ . The idea is that after mapping $H$ to $e$ , every pair of elements that differ by an element in $H$ now differ by $e$ , and are therefore the same element. Define $\sim$ so that $a\sim b$ whenever $a,b$ differ by an element in $H$ . Like before, we can express this relation more succinctly: $\begin{aligned} a\sim b\iff&a,b\text{ differ by some }h\in H\\ \iff&ah=b&\text{ for some }h\in H\\ \iff&h=a^{-1}b&\text{ for some }h\in H\\ \iff&a^{-1}b\in H\\ \iff&b\in aH \end{aligned}$ the idea being that $aH$ contains all elements that differ from $a$ by a factor of some $h\in H$ , and if $b$ is one of them, then $a,b$ differ by some $h$ as required.

Is $\sim$ a equivalence relation? Let’s see:

Reflexivity: $a\in aH$ since $H$ contains $e$ , and therefore $aH$ contains $ae=a$ .
Symmetry: Given $b\in aH$ , we know $a\in bH$ since $\begin{aligned} &b\in aH\\ \iff&a^{-1}b\in H\\ \iff&b^{-1}a\in H&\text{ since }H\text{ is closed under inverse}\\ \iff&a\in bH \end{aligned}$
Transitivity: Given $b\in aH$ and $c\in bH$ , we will show $c\in aH$ : $\begin{aligned} &c\in bH\text{ and }b\in aH\\ \iff&b^{-1}c\in H\text{ and }a^{-1}b\in H\\ \iff&(a^{-1}b)(b^{-1}c)\in H&\text{ since }H\text{ is closed under product}\\ \iff&a^{-1}c\in H\\ \iff&c\in aH\\ \end{aligned}$

Since $\sim$ is an equivalence relation, we can use it to partition $G$ into equivalence classes $[g]$ . Like before, we note that $b\in [a]$ iff $a\sim b$ iff $b\in aH$ , so each equivalence class $[a]$ is exactly $aH$ .

Now, do these equivalence classes form a group? Let’s see:

Here we encounter a problem: unlike the center

Z(G)

, the subgroup

H

doesn’t commute with all elements, and therefore

aHbH

is not equal to

abH

in general.

Contains identity: $[e]$ satisfies $[e][g]=[eg]=[g]=[ge]=[g][e]$ , and is therefore the identity.
Contains product: We need to show that the product $[a][b]$ yields another equivalence class. $[a][b]=(aH)(bH)=aHbH\ne abH=[ab]$

Recall that a subset $S$ of $G$ commutes with all elements of $G$ exactly when $S$ is invariant under conjugation. So we need the subgroup $H$ to be invariant under conjugation, and then we can proceed with showing that the equivalence classes $aH$ form a group by letting $H$ commute with every element.

Therefore: If

H\le G

is invariant under conjugation, then the equivalence classes of

G

under the relation

a\sim b

iff

b\in aH

form a group.

Contains identity: $[e]$ satisfies $[e][g]=[eg]=[g]=[ge]=[g][e]$ , and is therefore the identity.
Contains product: $[a][b]=(aH)(bH)=aHbH=abHH=abH=[ab]$ so the product of two equivalence classes $[a],[b]$ is $[ab]$ , which exists because $ab\in H$ by closure under product in $H$ .
Contains inverse: $[g]^{-1}=(gH)^{-1}=H^{-1}g^{-1}=g^{-1}H=[g^{-1}]$ so the inverse $[g^{-1}]$ of $[g]$ exists because $g^{-1}\in H$ by closure under inverse in $H$ .

Theorem: If a subgroup

H

has unique order, then it is normal.

First, note that the conjugate of a subgroup $gHg^{-1}$ is a subgroup of the same order:

Contains identity: $e=geg^{-1}\in gHg^{-1}$
Absorbs product: $(gHg^{-1})(gHg^{-1})=gHHg^{-1}=gHg^{-1}$
Absorbs inverse: $(gHg^{-1})^{-1}=gH^{-1}g^{-1}=gHg^{-1}$
Same order: the map $h\mapsto ghg^{-1}$ is a map $H\to gHg^{-1}$ and is a bijection since it has an inverse $h\mapsto g^{-1}hg$ . Thus $|H|=|gHg^{-1}|$ .

Since the conjugate of a subgroup must have the same order but $H$ has unique order, $H$ must be invariant under conjugation, thus normal.

When a subgroup $H\le G$ is invariant under conjugation, we call it a normal subgroup, and denote it by $H\lhd G$ . And as we’ve shown, sending the elements of $H$ to $e$ forms a quotient group $G/H$ if and only if $H$ is a normal subgroup. The elements of every quotient group, the equivalence classes $aH$ , are called cosets of $H$ . So the above result can be written as:

Theorem: $H\lhd G$ iff the cosets of $H$ form a quotient group $G/H$ .

This enshrines the importance of normal subgroups – they are exactly the subgroups $H$ that you can send to $e$ to construct a quotient group. We already know of such a subgroup: the center of a group $Z(G)$ is always a normal subgroup of $G$ . This makes sense since $Z(G)$ is made up of elements that commute with every element in $G$ , so it must be invariant under conjugation. In fact any element of the center must generate a normal subgroup:

Theorem: Central elements generate normal subgroups.

Given that central elements $z\in Z(G)$ are invariant under conjugation, it follows that any product of a central element with itself is also invariant under conjugation. This means the subgroup $\<z\>$ generated by $z$ is necessarily invariant under conjugation in $G$ , and therefore a normal subgroup of $G$ .

Given $H$ a subgroup, $aH$ is a coset even if $H$ is not a normal subgroup — being a normal subgroup just means the cosets form a group. Nevertheless, the fact that you can make cosets out of any subgroup leads directly to a very important theorem:

Lagrange’s Theorem: The order of every subgroup divides the order of the group.

For any subgroup $H\le G$ , consider its cosets $aH$ . There is a bijection between any two cosets $aH$ and $bH$ : left-multiply $aH$ by $ab^{-1}$ to get $bH$ . It’s a bijection because the inverse is left-multiplying by $ba^{-1}$ . The existence of a bijection between every coset implies that every coset has the same size, and in particular they are all the same size as $eH=H$ , which is $|H|$ .

Since cosets are equivalence classes, they form a partition of the group $G$ . But since this partitions $G$ into equally-sized equivalence classes of size $|H|$ , $|H|$ must divide $|G|$ .

Corollary: The order of

G/H

|G|/|H|

We showed earlier that the cosets partition $G$ . Since the cosets are all the same size, $|H|$ , the number of such cosets must be $|G|/|H|$ . But the cosets are exactly the elements of the quotient $G/H$ , so the order of $G/H$ is $|G|/|H|$ .

We proved earlier that conjugacy classes are subsets of $G$ that are invariant under conjugation. Conjugacy classes have a close relationship with normal subgroups:

Theorem: Normal subgroups are exactly subgroups that are a union of conjugacy classes.

It is easy to see that if a subgroup is a union of conjugacy classes, then it is normal, since union preserves invariance under conjugation.

To show the converse, let $H\lhd G$ be a normal subgroup of $G$ and consider an arbitrary element $h\in H$ . Since $H$ must be invariant under conjugation, conjugating $h$ must give another element in $H$ . That is, $H$ contains all conjugates of $h$ , which means $H$ contains the conjugacy class $[h]$ . Since this is true for arbitrary $h\in H$ , $H$ fully contains the conjugacy class of each of its elements, meaning $H$ must be a union of conjugacy classes.

This theorem gives us a method to determine whether a subgroup is normal beyond checking for invariance under conjugation. All we need to do is find the conjugacy classes — then you can obtain every normal subgroup of the group by taking unions of conjugacy classes, and checking which unions form a subgroup.

For instance:

Example:

\{e\}

Z(G)

, and

G

are always normal subgroups of

G

We already know these three are subgroups of $G$ . To prove that they are normal, note that they are all unions of conjugacy classes:

$e$ is central and therefore is only conjugate to itself, so it is in a singleton conjugacy class.
$Z(G)$ consists of central elements and by the same argument, is a union of singleton conjugacy classes.
$G$ trivially contains all the conjugacy classes in $G$ .

Since all three subgroups are unions of conjugacy classes, they must be normal.

In this section, we make a group abelian.

Is there a way to quotient a group $G/H$ such that the resulting group is abelian?

To do this, we first look at the equation $gh=hg$ . If we move everything to one side, we get $g^{-1}h^{-1}gh=e$ . This tells us that if $g^{-1}h^{-1}gh$ is the identity element $e$ , then $g$ and $h$ commute. The element $g^{-1}h^{-1}gh$ is called the commutator of $g$ and $h$ , and is written $[g,h]$ .

(Note that $ghg^{-1}h^{-1}$ is also a commutator of $g$ and $h$ . We want to stick to one definition, so let’s use the first one, $g^{-1}h^{-1}gh$ .)

What if the only commutator in the group is the identity $e$ ? That means no matter what $g$ and $h$ you pick, their commutator $[g,h]$ is $e$ and therefore they commute. So if the only commutator in the group is $e$ , the group is abelian.

The idea here is that if we can send all the commutators to $e$ , the resulting group will be abelian. But to do this, we need to make sure that the commutators form a normal subgroup $H\lhd G$ , so that the quotient $G/H$ sends $H$ to $e$ . Do the commutators form a normal subgroup?

The commutators almost form a subgroup. The identity element is a commutator. The inverse of a commutator $[g,h]^{-1}$ is $[h,g]$ , a commutator. But the group product of two commutators need not be a commutator.

We can solve this problem by having the commutators generate a subgroup, i.e. include all their products as part of the subgroup. Let $G'$ be the subgroup generated by all the commutators. We call $G'$ the commutator subgroup, and much later we will see why it is also called the derived subgroup. For now, let’s prove that it is normal:

Theorem: The commutator subgroup

G'

is a normal subgroup of

G

WTS the conjugate of arbitrary $[g,h]\in G'$ is also a commutator. We can prove this directly.

$\begin{aligned} &k[g,h]k^{-1}&\text{the conjugate of }[g,h]\\ =&k(ghg^{-1}h^{-1})k^{-1}&[g,h]\text{ is a commutator }ghg^{-1}h^{-1}\\ =&(kgk^{-1})(khk^{-1})(kg^{-1}k^{-1})(kh^{-1}k^{-1})&\text{distribute}\\ =&(kgk^{-1})(khk^{-1})(kgk^{-1})^{-1}(khk^{-1})^{-1}\\ =&[kgk^{-1},khk^{-1}] \end{aligned}$

The conjugate of an arbitrary commutator in $G'$ is a commutator, so $G'$ is invariant under conjugation, and therefore a normal subgroup.

Now if we take the quotient $G/G'$ , we send all the commutators to $e$ and are left with a group whose only commutator is $e$ , which is an abelian group. This quotient $G^{ab}\equiv G/G'$ is called the abelianization of $G$ .

This easily extends to quotienting by any normal subgroup containing $G'$ as well.

Theorem: A quotient

G/N

is abelian iff

N

includes

G'

( $\to$ ) If $G/N$ is abelian, its only commutator is $e$ , which means any commutator of $G$ was sent to $e$ after quotienting by $N$ , which means $N$ includes the commutators of $G$ .
( $\from$ ) If $N$ includes all the commutators of $G$ , it means $G/N$ sends all the commutators to $e$ , which means $G/N$ is abelian.

In this section, we review abelian and centerless groups.

Abelian groups and centerless groups are both subclasses of groups. Moreover, you can study the abelian part and the centerless part separately; simply quotient by the commutator subgroup $G'$ to get the abelian part, and quotient by the center $Z(G)$ to get the centerless part. Doing both gives you the trivial group $\{e\}$ , since it is the only group that is both abelian and centerless.

In this section, we learn some more ways to find normal subgroups.

So far, we’ve encountered a few normal subgroups that always exist for a group $G$ :

The trivial subgroup $\{e\}$ is always a normal subgroup.
The center $Z(G)$ is a normal subgroup.
The group $G$ is itself a normal subgroup of $G$ .
The commutator subgroup $G'$ is a normal subgroup.

We also found that a subgroup is normal iff it is a union of conjugacy classes, but that requires finding the conjugacy classes of a group. Are there any other shortcuts?

Here are a few other shortcuts:

Theorem: Every subgroup of an abelian group is normal.

Since $gh=hg$ implies $g=hgh^{-1}$ for every $g,h\in G$ , every element of an abelian group is invariant under conjugation, and therefore so is every subgroup. So every subgroup of an abelian group is normal.

Theorem: For finite groups, a subgroup with unique order is normal.

Lemma: In a finite group, the conjugate of a subgroup $gHg^{-1}$ is a subgroup with the same order.
- We prove closure under group product, an identity element, and inverses exist for $gHg^{-1}$ .
- Closure: The product of $gh_1g^{-1}$ and $gh_2g^{-1}$ is $g(h_1h_2)g^{-1}$ , so $gHg^{-1}$ is closed under group product.
- Identity: $geg^{-1}=e$ shows $H$ contains $e$ .
- Inverse: $(ghg^{-1})^{-1}=gh^{-1}g^{-1}$ shows $H$ has inverses.
- Therefore $gHg^{-1}$ is indeed a subgroup. To show that it has the same order, notice that product and inverse above are the same as in the original group, except with the $g,g^{-1}$ around them. This means all elements in the subgroup were merely renamed, which is a bijection $H\leftrightarrow gHg^{-1}$ .
- Since the group is finite and a bijection exists between the subgroup and its conjugate, they must have the same order.
Since $H$ has unique order, we necessarily have $gHg^{-1}=H$ , which by definition means $H$ is invariant under conjugation, i.e. normal.

Theorem: The product

HK

of two subgroups

H,K\lhd G

is a subgroup iff at least one of

H,K

is normal.

Let $h\in H$ and $k\in K$ , then we can show that $HK$ is a subgroup:

Product is preserved: $(hk)(h'k')=hk(k^{-1}h''k)k'=(hh'')(kk')\in HK$ . This uses the fact that $kh'=h''k$ , assuming that $H$ is normal. The same argument can be made with $kh'=h'k''$ when $K$ is normal.
Inverse is preserved: $(hk)^{-1}=k^{-1}h^{-1}=k^{-1}(kh'k^{-1})=h'k^{-1}\in HK$ .
Contains identity: Being subgroups, both $H,K$ contain $e\in G$ , thus $HK$ also contains $e$ .

Theorem: The product

HK

of two normal subgroups

H,K\lhd G

is normal.

Since $H,K$ are normal, $HK$ is a subgroup. To show $HK$ is normal, consider $g\in G$ . The conjugate of $hk$ by $g$ is $ghkg^{-1}$ , which is equal to $(ghg^{-1})(gkg^{-1})$ . Since $H$ and $K$ are themselves normal, this is equal to some element $h'k'\in HK$ , so $HK$ is invariant under conjugation.

Theorem: The intersection

H\cap K

of two normal subgroups

H,K\lhd G

is a normal subgroup of both.

We know that the intersection of two subgroups is a subgroup of both. To show that the subgrouping is a normal subgrouping, note that if all conjugates of an element in $H$ is in $H$ , and all conjugates of an element in $K$ is in $K$ , then all conjugates of an element in both $H$ and $K$ are in both $H$ and $K$ .

In this section, we measure the normality of a subgroup.

Recall that in the very beginning, we noted that the set of all elements that commute with every element is somewhere between “just the identity element” and “every element.”

Similarly, the set of all elements that commute with some $g$ (the centralizer of $g$ ) is somewhere between $\<g\>$ and “every element.”

We can apply a similar approach to subgroups regarding normality. Recall that a subgroup $H\le G$ is normal iff every element $g\in G$ commutes with it: $gH=Hg$ . That is to say, iff the set of elements $g\in G$ satisfying $gH=Hg$ consists of the entire group. Call this set the normalizer of $H$ , denoted $N_G(H)$ .

Theorem: Every subgroup

H

is a normal subgroup of its normalizer

N_G(H)

The normalizer $N_G(H)$ is the set of elements $g\in G$ satisfying $gH=Hg$ . But that is exactly the condition $gHg^{-1}=H$ , i.e. $H$ is invariant under conjugation under elements in the normalizer, i.e. $H$ is normal in $N_G(H)$ .

Then a subgroup $H\le G$ is normal in $G$ when its normalizer $N_G(H)$ is “every element,” i.e. equal to $G$ . That’s the maximum; what’s the minimum? The set of elements $g$ such that $gH=Hg$ must always include the elements $h\in H$ , since $hH=H=Hh$ . So at minimum, $N_G(H)=H$ meaning $H$ is self-normalizing, and at maximum, $N_G(H)=G$ i.e. $H\lhd G$ .

Note these parallels between the centralizer and the normalizer of a subgroup:

A subgroup $H\le G$ is central in its centralizer: $H\subseteq Z(C_G(H))$ . Adding more elements to the centralizer would make $H$ not central, so the centralizer is the ‘maximum’ superset of $H$ that makes $H$ central in it. If the centralizer of $H$ is $G$ , then $H$ is central in $G$ .
A subgroup $H\le G$ is normal in its normalizer: $H\lhd N_G(H)$ . Adding more elements to the normalizer would make $H$ not normal, so the normalizer is the ‘maximum’ superset of $H$ that makes $H$ normal in it. If the normalizer of $H$ is $G$ , then $H$ is normal in $G$ .

These statements hold true for arbitrary subsets $S\subseteq G$ as well (but replace “normal” with “invariant under conjugation.”) Note that again, the centralizer and normalizer are always subgroups of $G$ , even when $S$ is not a subgroup of $G$ .

Like the centralizer, the normalizer is always a subgroup:

Theorem: The normalizer

N_G(S)

is a subgroup of

G

Identity: Clearly $e\in N_G(S)$ because $eS=S=Se$ .
Inverses: If $g\in N_G(S)$ we have $gS=Sg$ , which is the same as saying $Sg^{-1}=g^{-1}S$ , therefore $g^{-1}\in N_G(S)$ .
Product: If $g,h\in N_G(S)$ , we have $gS=Sg$ and $hS=Sh$ . Another way to write that is $gSg^{-1}=S$ and $hSh^{-1}=S$ . Substituting, we get $ghSh^{-1}g^{-1}=S$ , which is equivalent to saying $(gh)S=S(gh)$ , thus $gh\in N_G(S)$ .
Since it has identity, inverses, and product, $N_G(S)$ is always a subgroup of $G$ .

Recall that to arrive at the class equation: $|G|=|Z(G)|+\sum_i|G|/|C_G(g_i)|$ we had to use the fact that every conjugacy class is in the form $bC_G(g)$ for some $b\in G$ . This was because the equality $aga^{-1}=bgb^{-1}$ can be rewritten as $(b^{-1}a)g=g(b^{-1}a)$ meaning two conjugates of $g$ are equal every time the element $b^{-1}a$ commutes with $g$ , and the set of these elements $b^{-1}a$ are the centralizer $C_G(g)$ .

How many conjugates of a subgroup $H$ are there? In other words, what is the size of the conjugacy class of $H$ ? In this case, we are looking at the equality $aHa^{-1}=bHb^{-1}$ which can be rewritten as $(b^{-1}a)H=H(b^{-1}a)$ meaning two conjugates of $H$ are equal every time the element $b^{-1}a$ commutes with $H$ . The set of these elements $b^{-1}a$ are exactly the normalizer $N_G(H)$ . So distinct conjugates of $H$ are in one-to-one correspondence with the cosets of the form $b^{-1}aN_G(H)$ . Thus by the same logic as before (all cosets are the same size), the number of conjugates of $H$ is equal to $|G|/|N_G(H)|$ . TODO make these both theorems

TODO this directly shows that if normalizer is group, then H is invariant under conjugation since there’s only one conjugate

By studying the normalizer, we get an alternate way of identifying normal subgroups. Here’s how.

Given a subgroup $H$ , consider the partition of $G$ into cosets that look like $gH$ . Also consider the partition of $G$ into cosets that look like $Hg$ . To distinguish the two, the cosets $gH$ are called left cosets and the cosets $Hg$ are called right cosets. Note that an element $g$ is in the normalizer of $H$ iff $gH=Hg$ , i.e. the left coset $gH$ coincides with the right coset $Hg$ . If this is true of $g$ , it must be true of every element in its coset $gH$ . Because of this, the normalizer of $H$ is a union of cosets. More precisely, $N_G(H)$ is exactly the union of cosets common to both partitions of $G$ .

One consquence of the normalizer having elements $g\in N_G(H)$ that satisfy $gH=Hg$ is that they are exactly the elements that make $H$ invariant over conjugation: $gHg^{-1}=H$ . In other words, the subgroup $H$ is normal in the normalizer: $H\lhd N_G(H)$ . This means that we can take the quotient $N_G(H)/H$ .

An interesting fact arises when $H$ is a $p$ -subgroup. Recall that $p$ -groups are groups of prime power order $p^n$ . Similarly, a $p$ -subgroup is a subgroup of prime power order $p^n$ .

Theorem: For every

p

-subgroup

H\le G

|N_G(H)/H|\equiv |G/H|\pmod p

Recall that in every quotient, like $N_G(H)/H$ , TODO left off here

the larger group $N_G(H)$ is partitioned into equally sized partitions with

Summary

We’ve learned that:

The center of a group $Z(G)$ is basically a measure of how “commutative” a group is. If the center consists of the whole group, the group is abelian (the most commutative). If the center is trivial, the group is the least commutative it can be.
Conjugacy classes are essentially subsets that commute. Subgroups formed by a union of conjugacy classes are commutative with the group and are called normal subgroups.
By quotienting a normal subgroup (sending all its elements to $e$ ) we can either make a group centerless (by quotienting by $Z(G)$ , the center of the group) or we can make the group abelian (by quotienting by $G'$ , the commutator subgroup).

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