I know there is a proof using these theorems:

  • The center of a finite p−group is non-trivial

  • For any group G , $G/Z(G)$ is cyclic iff $G$ is abelian, or in otherwords: the quotient $G/Z(G)$ can never be non-trivial cyclic.

But is there a proof not using these theorems?

  • $\begingroup$ The only proof that I can think of contains essentially the same arguments as you would use in proving the two theorems you mention, so it more efficient to prove these theorems independently. $\endgroup$ – Derek Holt Dec 4 '16 at 16:26

If there is an element of order $p^2$, it's cyclic and thus abelian. Suppose there is no element element of order $p^2$. Then, the order of of the elements of $G$ are either $1$ or $p$. Let $h_1,h_2\in G$ two elements of order $p$ s.t. $h_2\notin\left<h_1\right>$. Then, $\left<h_1,h_2\right>$ is of order $p^2$ and is s.t. $|\left<h_1,h_2\right>|\geq p+1$. Therefore, $|\left<h_1,h_2\right>|=p^2$, and thus $G=\left<h_1,h_2\right>$. Therefore, $G$ is abelian.

We can show that $\left<h_i\right>$ are normal in $G$. Then, $[G:H_i]=p$ and thus $G/H_i$ are cyclic, and thus abelian. Let consider $$\pi: G\longrightarrow G/H_i,$$ defined by $\pi(g)=gH_i$. Take an element of $[G,G]=\left<ghg^{-1}h^{-1}\mid g,h\in G\right>$. You have that $$\pi(ghg^{-1}h^{-1})=\pi(g)\pi(h)\pi(g^{-1})\pi(h^{-1})\underset{G/H_i\ cyclic}{=}H_i$$ and thus and thus $[G,G]\leq H_i$, and since $H_1\cap H_2=\{1\}$, we get $[G,G]=\{1\}$. Therefore $G=G/[G:G]$ is abelian.

  • 2
    $\begingroup$ Why does $h_1 h_2 = h_2 h_1$ ? $\endgroup$ – lhf Dec 4 '16 at 13:45
  • $\begingroup$ @lhf: I completed my proof. $\endgroup$ – Surb Dec 4 '16 at 14:00
  • $\begingroup$ But you haven't shown that the $\langle h_i \rangle$ are normal in $G$. $\endgroup$ – Derek Holt Dec 4 '16 at 16:24
  • $\begingroup$ @DerekHolt: I will not do all the work ;-) But I agree, it's not the easiest part. $\endgroup$ – Surb Dec 4 '16 at 17:30
  • $\begingroup$ btw $k_4$ is abelian not cyclic $\endgroup$ – RAM_3R Feb 25 at 22:20

Now for something completely different (using representation theory over $\mathbb{C}$)...

Let $G$ be a group of order $p^2$. Let $Irr(G)$ be its set of complex irreducible characters. Note that the principal character $1_G \in Irr(G)$. Since $|G|=p^2=\sum_{\chi \in Irr(G)}\chi(1)^2=1 + \sum_{\chi \in Irr(G)-\{1_G\}}\chi(1)^2$, and for all $\chi \in Irr(G)$: $\chi(1) \mid p^2$, it follows that all $\chi$ must be linear, which is equivalent to $G$ being abelian.


Sorry to revive a very old thread -- and if someone better versed in netiquette says so I'll just delete this -- but I was looking for a nice low-level proof of this fact (which is what brought me to this page in the first place) and was delighted to find two fairly straightforward and low-level proofs of this nice fact. Here's the slightly simpler one (basically thieved off the 'net somewhere, but I forget where); the most sophisticated fact here is Fermat's Little Theorem. I thought I'd record it in case anybody else is ever interested.

Let $G$ be a non-cyclic group of order $p^2$ with identity element $e$, and let $g,h\in G$ of order $p$ such that $\langle g\rangle\cap \langle h \rangle = \{e\}$. Assume by way of contradiction that $g$ and $h$ do not commute. I claim that $ghg^{-1}$ cannot be a power of $h^k$ of $h$, for if it were, then one calculates that $g^rhg^{-r} = h^{k^r}$ for all $r$, and then $g^{p-1}hg^{1-p} = h^{k^{p-1}}$, which is equal to $h$ by Fermat. This would imply $g^{-1}h=hg^{-1}$ and now $gh=hg$, contradiction.

So $\langle h\rangle$ and $\langle ghg^{-1}\rangle$ are distinct subgroups of order $p$, and the distinct cosets of $\langle ghg^{-1}\rangle$ in $G$ are $\langle ghg^{-1}\rangle, h\langle ghg^{-1}\rangle, h^2\langle ghg^{-1}\rangle, \dots, h^{p-1}\langle ghg^{-1}\rangle$. Therefore $g^{-1}$ lies in some $h^k\langle ghg^{-1}\rangle$, so $g^{-1}=h^kgh^{\ell}g^{-1}$ for some $\ell$. Then $e=h^kgh^{\ell}$ and $g=h^{-k-\ell}$, contradicting that $\langle g \rangle$ and $\langle h \rangle$ intersect trivially.

Well, now we are done: $g$ and $h$ commute, so the mapping from $\mathbb Z_p\times \mathbb Z_p$ to $G$ taking $(a,b)$ to $g^ah^b$ is an isomorphism.


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