# Proof by induction of Sylow's theorem

Prove directly that if $p$ is a prime and $p^{\alpha}\ | \ o(G)$, then $G$ has a subgroup of order $p^{\alpha}.$

How can I prove this, by induction on the order of the group $G,$ without using the existence of a $p$-Sylow subgroup?

• (One of) Sylow's theorem(s) is that assertion.
– Pedro
Nov 7, 2013 at 4:42
• @PedroTamaroff Yes it is but I am looking for a different proof without using the existence of a p-Sylow subgroup. Nov 7, 2013 at 4:49
• My point is that the existence of p-Sylow subgroups is that claim.
– Pedro
Nov 7, 2013 at 4:51
• @PedroTamaroff I see. My book gave three different proofs of Sylow's theorem so I got confused which proof this question was referring to. Nov 7, 2013 at 4:55
• @Pedro The usual version of Sylow asserts the existence of Sylows, not the existence of subgroups of all prime power orders. Naively, it is initially conceivable that one could use the soft version of Sylow in order to more quickly prove the strengthened version, so the special request makes some sense. (On the other hand, there is some reason to expect the same proof of soft Sylow can be augmented to do the strengthened version, so it is a bit off to disclaim the same proof methods.)
– anon
Nov 7, 2013 at 4:58

CLAIM Let $k\geqslant 0$, $p$ a prime and $p^k\mid |G|$. Then $G$ contains a subgroup of order $p^k$.

P If $|G|=1$, there is nothing to prove. So suppose $|G|>1$ and the theorem proven for every group of order less than that of $G$. It is a theorem of Cauchy that if $G$ is a finite (abelian) group and $p\mid |G|$ then $G$ contains an element of order $p$. Using this, consider the class equation $$|G|=|C|+\sum [G:C(x_i)]$$

If $p\not\mid |C|$ then $p\not\mid [G:C(x_i)]$ for some $i$. This means that $p^k\mid |C(x_i)|$ and $|C(x_i)|<|G|$ so we're done. If $p\mid |C|$, $C$ is abelian, so we have an element $g$ of order $p$, and the order of $G/\langle g\rangle$ (we can take the quotient since $g$ is central) is divisible by $p^{k-1}$. By the inductive hypothesis we have a subgroup of order $p^{k-1}$, of the form $H/\langle g\rangle$ where $\langle g\rangle \subseteq H\leq G$. But $$|H|=[H:\langle g\rangle]||\langle g\rangle |=p^k$$ and the theorem is proven.

In particular, if $G=p^nk$ with $(p,k)=1$, there exists a subgroup of order $p^n$, i.e. a Sylow $p$-subgroup.

To prove Cauchy's theorem, we can do something very similar. If $G$ is abelian, take an element $g\in G$. If this element has order divisible by $p$, say $=pr'$ then $g^{r'}$ is an element of order $p$. Else, consider $G/\langle g\rangle$ (we can quotient since $G$ is abelian). This has order still divisible by $p$ but smaller than that of $G$. We obtain an element $g'\langle g\rangle$ of order $p$. Let $f$ be the order of $g'$, then $(g'\langle g\rangle) ^f=g'^f\langle g\rangle=\langle g\rangle$ so the order $p$ of $g'\langle g\rangle$ divides $f$. But then $g'$ has order divisible by $p$ and the previous case finishes things off. The general case then follows in the same manner as above, with the class equation.

• @Pedro, Cauchy's Theorem's true for any finite group, not only abelian ones. Nov 7, 2013 at 4:58
• I think the notation here may be a little off the standard (?) one: For one , and since the class equation was explicitly mentioned, $\;C=Z(G)=$ the center of the group $\;G\;$ and $\;C_{x_i}=$ stabilizer of $\;x_i\;$ (under conjugation) . Nov 7, 2013 at 5:01
• @DonAntonio True. Note though that the class equation makes it simpler to prove: just consider the abelian case and do something similar with the class equation in the nonabelian case.
– Pedro
Nov 7, 2013 at 5:02
• @PedroTamaroff: You are using induction on both $k$ and $|G|$ simultaneously here. While it is still correct, one needs to check a few more base cases before proceeding. Oct 8, 2018 at 15:52
• @PrahladVaidyanathan Well, you can do induction on $k+|G|$.
– Pedro
Oct 8, 2018 at 16:04