2

I found this theorem in the second edition of Topics in Algebra by I.N. Herstein.

The statement of the theorem is: If $G$ is an abelian group of order $o(G)$, and if $p$ is a prime number, such that $p^\alpha|o(G)$, $p^{\alpha +1}\nmid o(G)$, then $G$ has a subgroup of order $p^\alpha$

I am just wondering whether the condition $p^{\alpha +1}\nmid o(G)$ is necessary or not because I am pretty sure I figured out a proof that doesn't require that fact.

The following is a "proof" of the theorem without the fact that $p^{\alpha +1}\nmid o(G)$

Proof:

We will prove this with induction on $\alpha$. This is clearly true for $\alpha =0$. Let's say this is true for $\alpha = n$. Now consider the case where $\alpha=n+1>0$. So let $G$ be an abelian group such that $p^{n+1}|o(G)$.

According to Cauchy theorem for abelian group (Suppose $G$ is a finite abelian group and $p|o(G)$, where $p$ is a prime number. Then there is an element $a\neq e\in G$ such that $a^p=e$) we can pick an element $a\in G$ such that $o(a)=p$. Consider $N$ the cyclic group generated by $a$, clearly $o(N)=p$. It follows that $p^n|o(G/N)$. Clearly $G/N$ is also an abelian group, so by induction there exist a subgroup in $G/N$ whose order is $p^n$.

Let the element of that subgroup be $H=\{Na_1,Na_2,...,Na_{p^n}\}$. Clearly for any of those elements $Na_i$ there are exactly $o(N)=p$ elements $t$ in $G$ such that $Nt=Na_i$. So, consider all elements $t$ in $G$ such that $Nt$ is in $H$, let this set be $S$. Clearly there are $p^{n+1}$ elements in $S$. We will prove that $S$ is a subgroup of $G$

It is sufficient to prove that $S$ is closed under multiplication since $S$ is finite. $\forall t,k\in S$, $Nt=Na_i$ and $Nk=Na_j$ for some $i,j$. So since $H$ is a group then $\exists m$ such that $N(tk)=NtNk=Na_iNa_j=Na_m$, hence $N(tk)=Na_m$ which means $tk\in S$. So $S$ is closed under multiplication.

So, $S$ is a subgroup of $G$ with $p^{n+1}$ elements. Hence by induction the theorem is proved.

  • 2
    I did not gheck your proof, but yes, a finite group (need not even be abelian) has subgroups of any prime power order that divides the groups order. The focus on maximal such is because these are the harder ones to show existence of (the rest follow easily) and they have some further nice properties. – Tobias Kildetoft Dec 22 '16 at 21:45
  • @TobiasKildetoft I see, thank you for letting me know. May I know the name of that result? –  Dec 22 '16 at 21:54
  • @DougM Since $G$ is abelian then it follows that $N$ is normal, hence the quotient group exist CMIIW. –  Dec 22 '16 at 21:55
  • 1
    I see my confusion. The Sylow theorem does not require $G$ to be an abelian group. if $p^m|o(G)$ and $p^{m+1}\nmid o(G)$ then $G$ has a subgroup of order $p^m$ Adding the additional restriction that $G$ is abelian, then indeed the $p^{m+1}\nmid o(G)$ criteria is not necessary. – Doug M Dec 22 '16 at 22:35
  • @DougM It also is not necessary when $G$ is not abelian. – Tobias Kildetoft Dec 23 '16 at 06:36

2 Answers2

0

Sylow proved this for all finite groups in his paper where he proved the Sylow theorems.

KCd
  • 46,062
0

Your proof is fine, the condition is indeed not necessary. Another approach would be to prove that an abelian group of order $p^n$ has a subgroup of order $p^m$ for every $m\leq n$. This can also be done by induction, if you like.

The Sylow theorem is not very interesting for abelian groups, in view of the structure theorem for finitely generated abelian groups. An interesting generalization is that the theorem holds for all finite groups.

Servaes
  • 63,261
  • 7
  • 75
  • 163