# Proving that a subgroup $|H|=p^k$ is a Sylow subgroup of $|G|=p^km$, $m\nmid p$

I'm attempting to prove Sylow's theorems following the sketch described in the Wikipedia article, but I've run into a little hitch since the theorems are presented in a few slightly different forms in the article. First, we have the version stated in the proof itself:

Sylow 1A: A finite group $G$ whose order $|G|$ is divisible by a prime power $p^k$ has a subgroup of order $p^k$.

I was able to follow this proof without issue. But this statement differs slightly from the version stated in the introduction:

Sylow 1B: For any prime factor $p$ with multiplicity $n$ of the order of a finite group $G$, there exists a Sylow $p$-subgroup of $G$, of order $p^n$.

If one uses the definition that a Sylow $p$-subgroup is a subgroup of order $p^n$ where $v_p(|G|)=n$, then this follows trivially from (1A), but the actual definition given for a Sylow $p$-subgroup is as a $p$-group which is not a proper subgroup of a $p$-subgroup of $G$. So we need to prove the equivalence of these two definitions. (I have already shown that a finite group is a $p$-group iff it is of the form $|G|=p^n$ for some $n$.)

Let $n=v_p(|G|)$. Obviously a subgroup of order $p^n$ is a Sylow subgroup, because it is a $p$-group and any proper superset that is a $p$-group must have order a multiple of $p^{n+1}\nmid |G|$ in violation of Lagrange's theorem, and conversely a Sylow subgroup must have order at most $p^n$ from Lagrange's theorem, so it suffices to consider a Sylow subgroup $H$ of order $p^k$ for $k<n$. Then Sylow 1A gives a $p$-group $K$ of order $p^n$, which is also a Sylow subgroup.

This is in contradiction to Sylow 2, which states that all Sylow subgroups are conjugate and hence isomorphic and equinumerous, but the proof of this seems to already require this equivalence of definitions. Specifically, the proof given shows:

Sylow 2A: If $H$ is a $p$-subgroup of $G$ and $P$ is a $p$-subgroup of $G$ with order $p^n$ where $n=v_p(|G|)$, then there exists an element $g$ in $G$ such that $g^{−1}Hg\subseteq P$.

And this statement is not very useful for proving that Sylow subgroups are equinumerous because $|H|\le |P|$ is already obvious (since $|H|=p^k$ for some $k\le n$). Is there a way to avoid this circularity of derivations?

Note: One theorem that would also patch this up is:

Any $p$-subgroup $|H|=p^k$ for $k<n$ is properly contained in a $p$-group $K\supsetneq H$.

• I'm not sure I understand your contradiction. The sentence "Then Sylow 1A gives a $p$-group $K$ of order $p^n$, which is also a Sylow subgroup" is true and as far as I can see doesn't contradict anything. If you had written "Then Sylow 1A gives a $p$-group $K$ of order $p^k$, which is also a Sylow subgroup", then the first part of the sentence is true (such a subgroup does exist), but the second is not (if $k<n$ we don't call the subgroups of order $p^k$ Sylow subgroups). – Bungo Jan 17 '15 at 0:55
• Your post is slightly hard to follow. Wikipedia's proof is not circular. – Pedro Tamaroff Jan 17 '15 at 0:56
• @PedroTamaroff The things which are actually proven are not circular, but there are some holes in the proof in going from one equivalent definition to the other without proving the equivalence. – Mario Carneiro Jan 17 '15 at 0:58
• @Bungo The issue is that I need to prove that a subgroup of order $p^k$ for $k<n$ is not a Sylow subgroup, using the definition involving proper subgroups. Some authors use the definition that a subgroup is Sylow iff if has order $p^n$, but I don't know this to be true from my definition yet (that's what I'm trying to prove). – Mario Carneiro Jan 17 '15 at 1:00
• @Bungo The contradiction I am referring to there is that I have constructed two Sylow subgroups of different cardinalities - these can't be conjugate to each other because conjugate groups are isomorphic. – Mario Carneiro Jan 17 '15 at 1:02

At first glance, it seems they are using the definition that a Sylow subgroup is one of cardinality $p^n$ where $G$ has order $p^nm$ and $(m,p)=1$. It is actually nontrivial to show the equivalence that $P$ is a maximal $p$-subgroup of $G$ iff it has cardinality $p^n$. Note that it is trivial to show any group $G$ contains maximal $p$-subgroups: in the finite case, there is nothing to do, since any chain of $p$-groups eventually stops to give a maximal $p$-group. If $G$ is infinite, use Zorn's lemma. Here I spell out the argument in the link.

• First, they show the existence of $p$-subgroups of order $p^k$ for any $k\leqslant \nu_p(|G|)$. (This backs up the claim they want Sylow subgroups to be those with $k$ largest.)
• Second, they prove a useful lemma: if $G$ is a finite group acting on a finite set $X$ and $X_0$ is the set of fixed points of the action, $|X|=|X_0|\mod p$.
• Third, they show any $p$-subgroup $H$ of $G$ can be conjugated by an appropriate element to make it sit inside a Sylow $p$-subgroup $P$, for any choice of $H$ and any choice of $P$. This uses only point one: we only need that one Sylow subgroup exists to carry out the argument.
• Finally, since conjugation preserves cardinality, this gives any two Sylow subgroups are conjugate. Since all is finite, $A\subseteq B$ and $|A|=|B|$ implies $A=B$. This is what they use. They do not use the definition that Sylow subgroups are maximal $p$-subgroups. In fact, this proves the equivalence, as I note below.

For a finite group $G$ of order $p^nm$, $(m,p)=1$, a subgroup of order $p^n$ and a maximal $p$-subgroup are equivalent objects: if $P$ is a $p$-subgroup of order $p^n$, and $H$ is a $p$-subgroup of order $p^k$ with $H\geqslant P$, then $k\leqslant n$ because $p^k\mid p^nm$ and $(m,p)=1$, and $n\leqslant k$ because $p^n\mid p^k$. Hence $P$ is maximal.

Suppose now that $P$ is a maximal $p$-subgroup. Let $Q$ be any $p$-Sylow subgroup of $G$. By definition, this is a subgroup of order $p^n$. By the third bulletpoint, there exists $g$ for which ${}^g P\leqslant Q$, i.e. $P\leqslant {}^{g^{-1}}Q$. Since conjugation preserves cardinality and $P$ is a maximal $p$-subgroup, $P$ must have order $p^n$.

Here I leave a way to fix Wikipedia's argument, i.e. avoid claiming directly that if $P$ is a maximal $p$-subgroup of a finite group $G$ then $|G:P|\neq 0\mod p$.

Let $G$ be a finite group, and let $P$ be a Sylow subgroup. Then $|N(P):P|$ is prime to $p$. Indeed, if it was not, there would be $a\in N(P)$ such that $\bar a=aP$ has order $p$ in $N(P)/P$ by Cauchy. By correspondence, there would exist $H$ over $P$ and under $N(P)$ such that $H/P=\langle \bar a \rangle$. But $|H|=|H:P||P|$ so $H$ is a $p$-group strictly over $P$, which is absurd. Hence $|N(P):P|$ is prime to $p$. Now suppose $H$ is a $p$-subgroup inside $N(P)$. Then I claim $H\leqslant P$. Indeed, $PH$ is a subgroup of $N(P)$ since $P$ is normal in $N(P)$, and $P\leqslant HP$. But $HP$ has order a power of $P$ i.e. it is a $p$-group so by maximality $PH=P$ and $H\leqslant P$.

Now take a Sylow $p$-subgroup $P$. Then $G$ acts by conjugation on the set of conjugates of $P$. If we restrict the action of $G$ to $P$, the lemma above implies the only orbit of size one is $\{P\}$: saying that $g\in P$ fixes a conjugate $hPh^{-1}$ means that $hPh^{-1}$ sits inside $N(P)$. By the lemma, $hPh^{-1}$ sits inside $P$, and by cardinality must equal $P$. This gives the number of Sylow subgroups is congruent to $=1\mod p$ by looking at the partition of the set of conjugates under the action. This also gives that the action is transitive, for if $Q$ is any other Sylow subgroup and we restrict the action to $Q$, the above congruence gives there is at least one orbit of size one, and the same argument shows that $Q$ must equal one of the conjugates.

We conclude that the number of Sylow subgroups $n_p$ is $|G:N(P)|$ for any Sylow $P$ and is congruent to $1\mod p$; in particular not divisible by $p$. But then $|G:P|=|G:N(P)||N(P):P|$ is not divisible by $p$ either. Hence $P$ is a "maximal $p$-power subgroup", as desired. But we had to prove Sylow's theorems all over to get here!

• In the proof of Sylow 2, it says "Now $p\nmid [G : P]$ by definition", but this is assuming that $|P|=p^n$, when I only know that $P$ is a Sylow subgroup (which entails the lack of proper $p$-group supersets). Showing that $|P|=p^n$ when $P$ is Sylow is the issue. – Mario Carneiro Jan 17 '15 at 1:07
• @MarioCarneiro You are correct. I would say that is a hole in the proof unless Wikipedia agrees to define a Sylow subgroups as a maximum $p$-power subgroup. – Pedro Tamaroff Jan 17 '15 at 2:39