Exploration 7: Sylow theory

We proceed by strong induction on $|G|$. The base case is simple: if $|G|=p$, then $G$ itself is our required subgroup of order $p^1$.

Otherwise, given that $p^k$ divides $|G|$, we must show that $G$ contains a subgroup of order $p^k$. The inductive hypothesis will let us claim that any group smaller than $G$ whose order is divisible by some prime power $p^j$, has a subgroup of order $p^j$.

First, recall that all groups satisfy the class equation: $$|G|=|Z(G)|+\sum_i|G|/|C_G(g_i)|$$ Given that $p^k$ divides the LHS, $|G|$, it must divide the RHS as well. Then $p$ divides the RHS, and by properties of divisibility, $p$ either divides both $|Z(G)|$ and the sum $\sum_i|G|/|C_G(g_i)|$, or it doesn’t divide either of them.

If $p$ divides $|Z(G)|$, then by Cauchy’s theorem, $Z(G)$ contains an element of order $p$. This element $a\in Z(G)$, being central, must generate a normal subgroup of $G$. That means we can consider the quotient $G/\<a\>$.

• Since $p^k$ divides $|G|$ and $|\<a\>|=p$, it follows that $p^{k-1}$ divides $|G/\<a\>|$. By inductive hypothesis, $G/\<a\>$ must contain a subgroup $H$ of order $p^{k-1}$.
• But since every coset in $G/\<a\>$ contains the same number of elements (namely $|\<a\>|=p$ elements) this means the union of the $p^{k-1}$ cosets that make up $H$ within $G$ form a subset of $G$ of size $p^k$.
• Since $H$ is a subgroup, this subset is the required subgroup of order $p^k$.

Otherwise, if $p$ fails to divide $\sum_i|G|/|C_G(g_i)|$, then at least one of the $|G|/|C_G(g_i)|$ in the sum must be indivisible by $p$.

• If $|G|$ is divisible by $p^k$ but $|G|/|C_G(g_i)|$ is indivisible by $p$, it follows that $|C_G(g_i)|$ contains all the factors $p$ of $|G|$ and is therefore divisible by $p^k$.
• Since the order of $C_G(g_i)$ is divisible by $p^k$, we’d like use the inductive hypothesis on $C_G(g_i)$ (because centralizers are always subgroups of $G$) to immediately show that it must contain a $p$-subgroup of order $p^k$, as required. But to use the inductive hypothesis, we need $|C_G(g_i)|<|G|$.
• We can show $|C_G(g_i)|<|G|$ by showing that $|G|/|C_G(g_i)|>1$. We show this by showing $|G|/|C_G(g_i)|=1$ is false — it would imply a singleton conjugacy class, but in the class equation, singleton conjugacy classes are counted in $|Z(G)|$, not the sum $\sum_i|G|/|C_G(g_i)|$. Thus $|G|/|C_G(g_i)|\ne 1$, so $|G|/|C_G(g_i)|>1$.

In summary, if there isn’t a central element of order $p$ that lets us inductively find a subgroup of order $p^k$, there must be a centralizer that contains a subgroup of order $p^k$.

By definition, no other $p$-subgroup properly contains a Sylow $p$-subgroup. Thus if a $p$-subgroup contains a Sylow $p$-subgroup, it must be improperly contained, i.e. the two subgroups are equal.

Now let $\Syl_p(G)$ denote the set of all Sylow $p$-subgroups of $G$, and consider the conjugation action of $G$ on $\Syl_p(G)$. That is to say, for every $g\in G$ and $P\in\Syl_p(G)$, $gPg^{-1}$ is another Sylow $p$-subgroup in $\Syl_p(G)$. Let $P,Q$ be two Sylow $p$-subgroups of $\Syl_p(G)$. Consider the $P$-orbit $O_P$ of $Q$ under this action (all the subgroups obtainable by conjugating $Q$ by elements of $P$). By the orbit-stabilizer theorem, we have: $$|O_P|=\frac{|P|}{|Stab_P(Q)|}$$ Here, $|O_P|$ can only be a power of $p$ since its factors come from the order of the $p$-subgroup $|P|=p^k$. Note that if $p^0=1$, that would imply the orbit is $\{Q\}$ and that $Q$ is invariant under conjugation by all elements $g\in P$. But by the lemma, if $Q$ is invariant under conjugation by an element $g$ of order $p^k$, then $g\in Q$. Since this is true for all $g\in P$, we have $P\subseteq Q$, and since subgroups containing Sylow $p$-subgroups are equal, $P=Q$. Therefore there is only one orbit of order $p^0=1$, and all the others have order a power of $p$.

Since conjugation by $g\in G$ is always a transitive group action, every Sylow $p$-subgroup in $\Syl_p(G)$ can be obtained by conjugating $Q$ by elements in $G$. Thus $\Syl_p(G)$ can be considered the union of all $P$-orbits of $Q$. Again, there is one orbit of order $1$ among all other orbits of order a power of $p$, so we have $|\Syl_p(G)|\equiv 1\bmod p$.

• First part:
  • Let $S$ be the set of all Sylow $p$-subgroups of $G$, which necessarily have the same order $p^n$ since they are maximal.
  • The conjugate of a subgroup is a subgroup of the same order. Then conjugating the Sylow $p$-subgroups in $S$ permutes that set.
  • Conjugating the subgroups in $S$ by $g$ will fix the Sylow $p$-subgroups that contain $g$, because subgroups are closed under group product.
  • Conjugating the subgroups in $S$ by $P$ will fix only $P$ itself (since all Sylow $p$-subgroups are maximal, only $P$ contains elements only in $P$.) This means all other Sylow $p$-subgroups will be permuted to another Sylow $p$-subgroup.
  • But then conjugating by $P$ permutes $S\setminus\{P\}$ with no fixed points.
  • Each element of $P$ generates a cyclic subgroup $H$, which by Lagrange’s Theorem, divides $|P|=p^n$ .
  • Furthermore, the cycle length $k$ of this permutation must divide the order of that cyclic subgroup, because conjugating by $H$ $k$ times must lead you back to the same subgroup. Therefore, every cycle in this permutation must be a power of $p$.
  • The fact that there are no fixed points implies that there are no cycles of length $1$, and therefore every cycle in the permutation is divisible by $p$.
  • If a permutation on a set has only cycles divisible by $p$, then the size of the set is divisible by $p$. This means $|S\setminus\{P\}|=|S|-1$ is divisible by $p$, which implies $|S|\equiv 1\text{ mod }p$.
• Second part:
  • Conjugation by an arbitrary element $g\in G$ is a permutation of $S$.
  • Define the homomorphism $f:G\to S_{|S|}$, which maps elements of $G$ to the corresponding permutation in the symmetric group of $S$, $S_{|S|}$.
  • The kernel $K$ of this homomorphism is a normal subgroup of $G$ containing all the elements of $g$ that fix $S$ via conjugation. (i.e. they lead to the identity permutation in $S_{|S|}$.) Then elements of $G/K$ represent all the distinct permutations of $S$ by conjugation of some element in $G$.
  • But since there are exactly $|S|$ distinct Sylow $p$-subgroups reachable by such permutations, the number of distinct permutations divide $|S|$, so $|G/K|$ divides $|S|$.
  • Since $|G|=|G/K||K|$, we know that $|G|=(k|S|)|K|$ for some $k$. We can rearrange this into $|K|=\frac{|G|}{k|S|}$ or $|K|=\frac{p^nq}{k|S|}$.
  • Since $|S|\equiv 1\text{ mod }p$, we know $|S|$ doesn’t contain any factor $p$, therefore $|S|$ divides $q$.

• Assume that $p>q$. (If $p=q$ then it’s a prime power order, therefore solvable.)
• Third Sylow Theorem: The number of Sylow $p$-subgroups $n_p$ is congruent to $1\text{ mod }p$. If the group is order $p^n q$, then $n_p$ divides $q$.
• In this case, since $n_p$ divides the prime $q$ it must be either $1$ or $q$. It cannot be $q$ because $n_p\equiv 1(\text{ mod } p)$ but $p>q$. Therefore $n_p=1$.
• Let $G_1$ be that single Sylow $p$-subgroup. Lagrange’s Theorem says the order of every subgroup divides the order of the group $pq$. We know that $p$-subgroups have prime power order $p^k$, so the order of $G_1$ must be $p$.
• $G_1$ is the only $p$-subgroup, so it is the only order $p$ subgroup. Recall that subgroups with unique order are normal. Therefore $G_1\lhd G$.
• Since $p$-groups like $G_1$ are solvable, $G_1\lhd G$ indicates that $G$ is solvable.

• Third Sylow Theorem: The number of Sylow $p$-subgroups $n_p$ is congruent to $1\text{ mod }p$. If the group is order $p^n q$, then $n_p$ divides $q$.
• Since $n_p$ divides $q^m$ it must be a power of $q$ between $1$ and $q^m$.
• But $n_p$ is congruent to $1\text{ mod }p$.
• Since none of $q,q^2,\ldots,q^n$ are congruent to $1\text{ mod }p$, $n_p$ must be $1$ – there is only one Sylow $p$-subgroup, $P$.
• Recall that subgroups with unique order are normal. Therefore $P$ is normal. Since $P$ has prime power $p^n$ it is solvable. So we have $P\lhd G$ and therefore $G$ is solvable.

• TODO

• Every integer between $1$ and $59$ is either prime, a prime power, or the product of two primes, except for the following:

  ┌──┬─────────┐
  │# │primes   │
  ├──┼─────────┤
  │12│2 2 3    │
  │18│2 3 3    │
  │20│2 2 5    │
  │24│2 2 2 3  │
  │28│2 2 7    │
  │30│2 3 5    │
  │36│2 2 3 3  │
  │40│2 2 2 5  │
  │42│2 3 7    │
  │44│2 2 11   │
  │45│3 3 5    │
  │48│2 2 2 2 3│
  │50│2 5 5    │
  │52│2 2 13   │
  │54│2 3 3 3  │
  │56│2 2 2 7  │
  └──┴─────────┘

• Most of these are of order $p^nq^m$ for prime $p,q$, and so we need only prove either $p,p^2,\ldots p^n\not\equiv 1\text{ mod }q$ or $p,p^2,\ldots p^n\not\equiv 1\text{ mod }q$ to show that they are solvable:

  ┌──┬─────────┬────────────────────────────┐
  │# │primes   │                            │
  ├──┼─────────┼────────────────────────────┤
  │12│2 2 3    │ 3 ≢ 1 (mod 4)              │
  │18│2 3 3    │ 2 ≢ 1 (mod 3)              │
  │20│2 2 5    │ 2,4 ≢ 1 (mod 5)            │
  │24│2 2 2 3  │ ?                          │
  │28│2 2 7    │ 2,4 ≢ 1 (mod 7)            │
  │30│2 3 5    │ --                         │
  │36│2 2 3 3  │ ?                          │
  │40│2 2 2 5  │ 2,4,8 ≢ 1 (mod 5)          │
  │42│2 3 7    │ --                         │
  │44│2 2 11   │ 2,4 ≢ 1 (mod 11)           │
  │45│3 3 5    │ 3,9 ≢ 1 (mod 5)            │
  │48│2 2 2 2 3│ ?                          │
  │50│2 5 5    │ 2 ≢ 1 (mod 5)              │
  │52│2 2 13   │ 2,4 ≢ 1 (mod 13)           │
  │54│2 3 3 3  │ 2 ≢ 1 (mod 3)              │
  │56│2 2 2 7  │ ?                          │
  └──┴─────────┴────────────────────────────┘

• TODO