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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 ?
1Spectre1
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  • 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 $p^{n-1}$. – S.C.B. Jan 15 '18 at 15:11
  • Do you maybe mean "there are no non-abelian simple groups of order less than $60$?" – lulu Jan 15 '18 at 15:17
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    @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. – 1Spectre1 Jan 15 '18 at 15:22
  • @S.C.B. Thank you. But I am not familiar with group action. You mean conjugate action on set. – 1Spectre1 Jan 15 '18 at 15:25
  • Not following. Obviously there are non-abelian groups of order less than $60$ so, as stated, your example is wrong. Please correct it. – lulu Jan 15 '18 at 15:26
  • @1Spectre1 I'm not talking about that, unfortunately. – S.C.B. Jan 15 '18 at 15:26
  • @1Spectre1 Actually this is better. https://math.stackexchange.com/questions/64371/showing-non-cyclic-group-with-p2-elements-is-abelian – S.C.B. Jan 15 '18 at 15:26
  • @lulu, I edited my question. I need easy argument. – 1Spectre1 Jan 15 '18 at 15:33
  • No group of order $p^2$ is simple. There is always a subgroup of order $p$. – lulu Jan 15 '18 at 15:34
  • How can you say this fact? $p|p^2$. Then, $G$ has any element of $a$ such that $|a|=p$ due to Cauchy theorem. Then $\langle a\rangle$ is normal subgroup because of commutativity. – 1Spectre1 Jan 15 '18 at 15:40
  • @S.C.B. I understand your comment:there are no simple groups of order $p^n$ where $n\geq 2$. If group non-abelian then center is non-trivial proper normal subgroup. If group abelian then there exist any element of $a$ order $p$( Cauchy Theorem). Then $\langle a\rangle$ must be proper normal subgroup order $p$ because group is abelian. Am I correct? – 1Spectre1 Jan 15 '18 at 15:53

2 Answers2

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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.

Port
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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.

Sunny
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