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I am to trying to prove this theorem: A finite $p$-group cannot be simple unless it has order $p$.

I have this:

Let $G = P$ and $|G|=p$; then there exists $N$, a normal subgroup of $G$ by Lagrange's theorem, such that $|N| \mid |G|$ so, $|N|= |e| = 1$ or $|N| = p$

If $|N|=p$ so $N=G$.

But I don't know how I can prove the part in the case of $|N|=|e| = 1$

Is my start correct?

Ops
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    Hint: The center of a group is always a normal subgroup. The center of any $p$-group is nontrivial. – Ethan Alwaise Nov 13 '16 at 22:20
  • I did my best to format your post; please let me know if this is what you wanted to post. – amWhy Nov 13 '16 at 22:20
  • Yes, that is. Thanks you. – Ops Nov 15 '16 at 14:28
  • @EthanAlwaise I know that, but this is my argument to solve the case if the order of N is 1? – Ops Nov 15 '16 at 14:30
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    Your post says you want to prove that a $p$-group $G$ of order not $p$ is not simple. For this you need only produce a normal subgroup $N$ that is not $G$ or trivial. I'm saying you can just take the center of $G$. – Ethan Alwaise Nov 15 '16 at 18:48

2 Answers2

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Suppose $|G|=p^n$ for $n>1$. Recall the Class Equation: $$ |G|= |Z(G)| + \sum_{g_1,\ldots,g_r} |G \colon C_G(g_i)|, $$ where $Z(G)$ is the center of $G$, $C_G(g_i)$ is the centralizer of $g_i$, and $g_1,\ldots,g_r$ are representatives for the distinct conjugacy classes of $G$ with more than one element (for otherwise it is in the center). The centralizer of $g_i$, $C_G(g_i)$, is a subgroup of $G$ so that by Lagrange's Theorem its order must divide $|G|$.

Since $|G|=p^n$, $|C_G(g_i)|=p^k$ for some $k<n$ (it cannot be that $k=n$ for then $C_G(g_i)=G$ and then $g_i \in Z(G)$). Then $|G \colon C_G(g_i)|= \frac{|G|}{|C_G(g_i)|}=p^{n-k}$. Now we have $|G| - \sum\limits_{i = 1}^r |G \colon C_G(g_i)| = |Z(G)|$. Since $p$ divides the left side, we must have $p \mid |Z(G)|$. In particular, $Z(G)$ is non-trivial. But $Z(G)$ is always a normal subgroup so that $G$ cannot be simple.

Note in the case of $n=1$, then $G$ must be isomorphic to $\mathbb{Z}/p\mathbb{Z}$, which is simple.

Daniel Fischer
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mathematics2x2life
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I believe there is a small problem with the final part of the solution provided by mathematics2x2life. The main part does include proving that $Z(G)$ is non-trivial. It is also true that $Z(G)$ is always a normal subgroup. However, one should also explicitly mention the case where $Z(G)=G$, in which the fact that $Z(G)$ is a normal subgroup doesn't help us. In that case, $G$ would be Abelian, and there is a separate theorem stating that every simple Abelian group is isomorphic to $\mathbb{Z}_p$ for some prime $p$.

Dean Gurvitz
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    Cauchy's theorem (if $p$ divides $|G|$ then $G$ has an element of order $p$) with the fact that every subgroup of an abelian group is normal. – egreg Sep 12 '18 at 11:05
  • Yes. Specifically, the subgroup generated by the element of order $p$ is what provides the contradiction – Dean Gurvitz Sep 12 '18 at 12:14