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It is known that if $G$ is a finite group of order $p^2$ , where $p$ is prime , then the center of $G$ is non-trivial i.e. $Z(G) \ne${$e$} , $e$ being the identity element of $G$. Can this be proved without using class equation ?

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    http://math.stackexchange.com/a/64978/583 – Jack Schmidt May 26 '14 at 05:23
  • @JackSchmidt Do you feel this should be marked as a duplicate, or just mentioned as related? (since the answer there is very nice for this question, but the question linked does not itself specify not using the class equation). – Tobias Kildetoft May 26 '14 at 08:23
  • @Tobias: Since my duplicate votes are binding, I try not to use them on borderline cases. This could be "different questions with the same answer", which I think are not supposed to be closed as duplicates. However, Arturo's answer is exactly the answer needed, so it seems silly to leave this question on the unanswered queue. – Jack Schmidt May 26 '14 at 09:32
  • I have added an answer similar to the one by Arturo. I also considered adding a more advanced one, using knit products, but the best reference I could find for this was http://ahmetsinancevik.com/pdfler/A12.pdf which is frankly too poorly formulated to be useful for this (I think Proposition 3.1 should give this, but I am not actually sure how to interpret what it says). – Tobias Kildetoft May 27 '14 at 08:24

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To have this question not remain unanswered, here is a variation of the answer given by Arturo Magidin at Showing non-cyclic group with $p^2$ elements is Abelian

First, we can assume that the group is not cyclic, so all non-identoty elements have order $p$ and thus the subgroup generated by such an element has order $p$ and index $p$.
Since $p$ is the smallest prime diving the order of the group, any subgroup of index $p$ is normal (see Normal subgroup of prime index but beware that the proof by myself there would turn this entire thing into a cyclic argument).

Now, pick two distinct such subgroups (so pick an element $h\neq 1$ and $k\not\in\langle h\rangle$). Now $H = \langle h\rangle$ and $K = \langle k\rangle$ are normal subgroups of order $p$ intersecting trivially, so we have $G = HK$ and in fact $G = H\times K$ so $G$ is abelian.

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