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Showing non-cyclic group with $p^2$ elements is Abelian

"Let $p$ be a prime number. Prove that any group $G$ of order $p^2$ is abelian. You may assume the fact that the centre of a $p$-group is non-trivial".

I understand from the question is that the group $G$ is a $p$-group, with $p^2$ number of elements in it (meaning it is finite). Also, the centre, i.e the set of elements where $g_1g_2 = g_2g_1$ for all $g_1, g_2 \in G$, is non-trivial, meaning $\neq e$.

So, the non-trivial bit shows that the center, $Z$, either generates the group, so it's cyclic and therefore abelian by definition. Or, it generates a subgroup of order p, and so $Z$ with some elements missing from group $G$ generate G, but as these groups commute (because they have a centre of $Z$), we can say the group is abelian.

Is this correct?

Community
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Kaish
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2 Answers2

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The center of a p-group can't be trivial. If $Z(G)=p^2$ then we are done. If $Z(G)=p$ then $G/Z(G)$ has order p and is cyclic. Thus, G is abelian using the G/Z theorem.

Amr
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  • Why can't it be trivial? Why are we done if $Z(G) = p^2$? Is $Z(G)$ the order of the centre or the centre of G? How do you get the second claim? – Kaish Dec 08 '12 at 18:53
  • If $Z(G)=p^2$, then $Z(G)=G$ – Amr Dec 08 '12 at 18:55
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    Whats G/Z theorem? Lagrange Theorem? – Kaish Dec 08 '12 at 19:08
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    $G/Z(G)$ is cyclic implies that $G$ is abelian. –  Dec 08 '12 at 19:33
  • Adding one detail: if $Z(G)=p$, then $|G/Z(G)|=p$. It is easy to check $Z(G)$ is a subgroup of $G$. By Lagrange theorem, $$\begin{align}|G|&=|Z(G)|[G:Z(G)]\ &= |Z(G)| |G/Z(G)| \ &= p |G/Z(G)| \ &=p^2\end{align}.$$ Which implies $|G/Z(G)|=p$. A group with prime order is cyclic. – user264745 Sep 12 '23 at 18:31
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This is similar to the answer above, but perhaps more complete. By the class equation, we know that the center (the commutative part) can't be trivial since the identity element is in it, and the order of it must be divisible by $p$. Now, as above, $G/Z(G)$ is cyclic, then this is $G/Z=\langle aZ\rangle$. That tells us $x=a^iz_1$ and $y=a^nz_2$ for any $x,y\in G$. Then $$xy=a^iz_1a^nz_2=a^nz_2a^iz_1=yx$$ since elements in the same cyclic group commute and elements in $Z(G)$ commute. Thus we have that the group is commutative.

Clayton
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