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I wonder $p$-groups of order $p^2$ simple or not. I know all of them are abelian. So center equal whole group.Hence, it is not useful to test simplicity.

  • How we can classify them according to whether they are simple or not. For example;

Let $G$ be a group such that $|G|=25=5^2$. Then $G$ has Sylow $5$-subgroup(s) order $25$ such that $n_5\equiv 1$ (mod $5)$.

  • I could not imagine how I would continue i.e., is it simple or not ?
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  • $\begingroup$ This can be generalized to the fact that there are no simple groups of order $p^{n}$ where $n \ge 2$. I suggest you prove this by considering the group action of the group $G$ on the set of subsets(note, not subgroup) of $G$ which are of cardinality $p^{n-1}$. $\endgroup$ – S.C.B. Jan 15 '18 at 15:11
  • $\begingroup$ Do you maybe mean "there are no non-abelian simple groups of order less than $60$?" $\endgroup$ – lulu Jan 15 '18 at 15:17
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    $\begingroup$ @lulu I dont mean it. If you click to link then you can see this: If $G=p^n$ for a prime number $p$ and a natural number $n$, then $Z(G)$ is nontrivial. To me, it is not useful argument for groups of order $p^2$. Because group is abelian. Actullay, this argument is not useful for all abelian group. $\endgroup$ – 1Spectre1 Jan 15 '18 at 15:22
  • $\begingroup$ @S.C.B. Thank you. But I am not familiar with group action. You mean conjugate action on set. $\endgroup$ – 1Spectre1 Jan 15 '18 at 15:25
  • $\begingroup$ Not following. Obviously there are non-abelian groups of order less than $60$ so, as stated, your example is wrong. Please correct it. $\endgroup$ – lulu Jan 15 '18 at 15:26
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The abelian groups of order $p^2$ are $\mathbb{Z}/p^2\mathbb{Z}, \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z} $ and both are not simple.

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  • $\begingroup$ why $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$ is not simple? $\endgroup$ – 1Spectre1 Jan 16 '18 at 14:11
  • $\begingroup$ It has a normal subgroup $\mathbb {Z} /p\mathbb {Z} \times {1} $ $\endgroup$ – Port Jan 16 '18 at 15:30
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Note that if p is the smallest prime dividing the order of G, then any subgroup of index p is normal, and since group of prime power order have subgroup of index p, hence they can not be simple.

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