Exploration 3: Products and quotients

According to Lagrange’s Theorem, the order of a subgroup divides the order of its group. But if the order of the group is a prime $p$, then the only divisors of $p$ are $1$ and $p$, implying that the only subgroups of $G$ are $1$ and $G$. Therefore $G$ is simple.

For abelian groups, every element is in the center, so every element is in its own conjugacy class, which means every subgroup is composed of conjugacy classes, therefore normal.

Recall the Internal Direct Product Theorem proved earlier, which requires us to find normal subgroups $H,K\lhd G$ with trivial intersection where $G=HK$ and $|G|=|H||K|$. Then $G\iso H\times K$.

Since $G$ is abelian, all subgroups are normal, so we just need to find subgroups that satisfy this property.

Given that we’re looking for $G\iso\<g\>\times K$, we must have $H=\<g\>$, where we choose $g$ to be an element of maximal order in $G$, so that no other element generates $g$. This constrains our possible $K$ in a few ways:

• Since we need $H\cap K=\{e\}$, no non-identity element in $K$ can have $g$ as a factor, otherwise it would be in $H=\<g\>$.
• Since we need $G=HK$, $K$ needs to be a complement to $H$ in the sense that every $g\in G$ is equal to some product $hk$ for $h\in H,k\in K$.
• Since we need $|G|=|H||K|$, that product must be unique.

Because $G$ is a $p$-group, we may take $G=\<g\>K$ where $K\iso G/\<g\>$. Intuitively, $K$ is the set of representatives in $G$ of every coset in $G/\<g\>$. (TODO prove).

The intersection $\<g\>\cap K$ is trivial because any element of $\<g\>$ maps to the identity coset in $G/\<g\>$, and any member of $K$ is a representative for some coset in $G/\<g\>$. Therefore, any element in both $\<g\>$ and $K$ must be a representative of the identity coset in $G/\<g\>$, which can only be $e$. Thus the intersection is trivial.

Now we prove that $G=HK$. Since

TODO left off here

Use the lemma to decompose $G$ as a direct product of $p$-groups. Then to each $p$-group apply the following:

From the previous proof, for any element $g\in G$ with maximal order in $G$, we have $G\iso\<g\>\times H$, where $\<g\>$ is cyclic by definition and $H$ is isomorphic to a direct product of cyclic groups by induction.

• First, view $F$ as an additive group, where $+$ is the group product and multiplication by a factor is group exponent.
• One way to prove a group is free is to prove that there’s a basis. We can prove that a minimal set of generators $X=\{x_1,x_2,\ldots,x_n\}$ for the group is a basis if we can only get the identity element $e$ via $c_1x_1+c_2x_2+\ldots+c_nx_n=0$ by setting all the $c_i$ to $0$. Since our group is finitely generated, a minimal set of generators $X$ exists.
• Focus on $c_1x_1+c_2x_2$ and replace it with $c_1(x_1+kx_2)+(c_2-kc_1)x_2$.
• Note that the new basis element $x_1+kx_2$ is simply $x_1$ shifted by some multiple of $x_2$, and since $X$ is a minimal basis, $x_1+kx_2$ should not be equal to any other basis element.
• However, looking at the coefficient of $x_2$ this means we can subtract multiples of any coefficient from any coefficient by shifting the basis elements. If we keep doing this between $c_1$ and $c_2$, then by Euclid’s algorithm we eventually end up with $c_1=c_2=\pm gcd(c_1,c_2)$.
• If we do this for all nonzero coefficients we get $\pm gcd(c_1,c_2,\ldots,c_n)$.
• Since $F$ is torsion-free, it has no (non-identity) elements of finite order, so there are only two cases. If every $c_i$ is $0$, then $gcd(c_1,c_2,\ldots,c_n)=0$. Otherwise, we can assume $gcd(c_1,c_2,\ldots,c_n)=1$, since we note that $c_1x_1+c_2x_2+\ldots+c_nx_n=0$ is still true if we divide all $c_i$ by their GCD.
• But (assuming $c_1$ is one of the nonzero coefficients) if $c_1=\pm 1$ in the equation $c_1x_1+c_2x_2+\ldots+c_nx_n=0$, then we can write $c_2x_2+\ldots+c_nx_n=-c_1x_1=\pm x_1$, meaning that $x_1$ is actually a linear combination of other basis elements. This means in the case that not all $c_i$ are $0$, $X$ is actually not a minimal set.
• Therefore the only solution to $c_1x_1+c_2x_2+\ldots+c_nx_n=0$ is if all $c_i$ are 0, thus $X$ is a basis for the free group $F$.

• Since $F$ is a torsion-free abelian group, it is free, and has a basis $X$. Let $n$ be the size of $X$.
• Since $F$ is free, you can write every element as a linear combination $c_1x_1+c_2x_2+\ldots+c_nx_n$ with $c_i\in\ZZ$ and $x_i\in X$.
• This is equivalent to a tuple $(c_1,c_2,\ldots,c_n)\in\ZZ^n$. Since this represents all elements of $\ZZ^n$ as well, you can define an isomorphism between the two.
• This means $F$ is isomorphic to $\ZZ^n$, which is by definition a direct sum of the infinite cyclic group $\ZZ$.

• This is a result of the previous theorems. If you split $G$ into a torsion subgroup $T(G)$ and a torsion-free part $G/T(G)$, then the torsion subgroup factors into a direct product of cyclic groups of prime power order (proof) and the torsion-free part factors into a direct sum of copies of $\ZZ$ (proof).
• It remains to show that $T(G)\times G/T(G)\iso G$.
• Let $K$ contain the torsion-free elements of $G$ ($=\{e\}\cup(G-T(G))$). Since $T(G)$ is torsion and $G/T(G)$ is torsion-free, you can uniquely write any element of $G$ as a product of their torsion and torsion-free components $tk$ where $t\in T(G)$ and $k\in K$. This is because in an abelian group, any two ways of writing $g=tk$ result in the same $t$ and $k$ being chosen: if $g=t_1k_1=t_2k_2$, then $t_1t_2^{-1}=k_1^{-1}k_2$ and the only element shared between the torsion LHS and the torsion-free RHS is the identity $e$, so $t_1t_2^{-1}=k_1^{-1}k_2=e$ and therefore $t_1=t_2$ and $k_1=k_2$.
• Since every element of $g$ is represented by the pair $(t,k)$ there is a bijection between $G$ and $T(G)\times K$. Since $K$ contains all the torsion-free elements of $G$, it is isomorphic to the torsion-free part $G/T(G)$ and therefore $G\iso T(G)\times G/T(G)$.

Abelian groups $G$ have a trivial derived subgroup, so we immediately have $\{e\}\lhd G$.

• ($\to$)
  • If $G_1$ is solvable, then by definition it has a derived series $\{e\}\lhd\ldots\lhd G^{(2)}\lhd G^{(1)}\lhd G$.
  • We know that quotienting by a derived subgroup gives an abelianization, therefore each quotient is abelian.
• ($\from$)
  • If every quotient $G_n/G_{n+1}$ in the series is abelian, then it implies $G_{n+1}$ includes $G_n$’s derived subgroup $G_n^{(1)}$. (Proof)
  • The derived subgroup is always a normal subgroup, so we have $G_n^{(1)}\lhd G_{n+1}\lhd G_n$.
  • Normal subgrouping is transitive, so $G_n^{(1)}\lhd G_n$.
  • Since this is true for every $G_n$, there is a derived series $\{e\}\lhd\ldots\lhd G^{(2)}\lhd G^{(1)}\lhd G$. Therefore $G$ is solvable.

• The order of an element must divide the order of the group (Lagrange’s Theorem).
• Since a group with prime order $p$ is only divisible by $1$ and $p$, the order of every non-identity element must be $p$.
• That means the group is cyclic, therefore abelian, therefore solvable.