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Prove or disprove a group with order $p^3$ is abelian if its has a normal subgroup of order $p^2$, where $p$ is an odd prime.

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Thanks to you all! Alex's usage of group actions reminded me about a useful result that I encountered before: Let $H\leq G$ and |G:H|=n<\infty then there exists a normal subgroup $N$ of $G$ such that $N\leq H$ and $|G:N|$ is a factor of $n$! So according to this fact one can show in the case when $|G:H|=p$ and p is the smallest prime dividing |G|, then |H|=|N| thus (N=)H is normal in G. Therefore every subgroup of G with order $p^2$ is normal.

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    I think this theorem is often referred to as the "$n!$ theorem" (n-factorial theorem).2018-03-08
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Hmm, I have a problem in my text that says ALL groups of prime power order have normal subgroups of every prime power order less than or equal to the order of your group. So, if having a normal subgroup of index $p$ guaranteed a group of order $p^3$ was abelian, you would have all groups of order $p^3$ abelian, and this is not the case.

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    This is correct, I made a mistake. Good job! But, you're statement of your theorem is slightly off. see my edit2012-04-22
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In any finite $p$-group, a maximal subgroup is normal; there are many ways of proving this (in a finite $p$-group, if $H$ is a proper subgroup of $G$, then $N_G(H)$ strictly contains $H$; if $K$ is any group and $H$ is a subgroup such that $[K:H]$ is the smallest prime that divides $|K|$, then $H$ is normal; etc). And a group of order $p^n$ has subgroups of order $p^i$ for all $i$, $0\leq i\leq n$. So any group of order $p^3$ has subgroups of order $p^2$, and they are always normal.

So the problem is tantamount to asking whether all groups of order $p^3$ are abelian. And the answer is "no"; the simplest way of showing this is to exhibit a group of order $p^3$ that is not abelian.

Up to isomorphism, there are two such for each odd prime $p$; probably the simplest to consider is the group of all $3\times 3$ matrices with coefficients in $\mathbb{Z}/p\mathbb{Z}$ of the following form: $\left(\begin{array}{ccc} 1 & a & c\\ 0 &1 & b\\ 0 & 0 & 1 \end{array}\right);$ let $M(a,b,c)$ be such a matrix; by computing the products explicitly, one can show that $M(a,b,c)M(x,y,z) = M(a+x, b+y, c+z+ay).$ It is then straightforward to verify that this is a group of order $p^3$, in which every element has order $p$, and that is not abelian, since $M(1,0,0)M(0,1,0) = M(1,1,1)$, but $M(0,1,0)M(1,0,0) = M(1,1,0)$.

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    @awllower: [here's a link to a proof](http://planetmath.org/?op=getobj&from=objects&name=NormalityOfSubgroupsOfPrimeIndex) in PlanetMath; I've been unsuccessful in locating it here.2012-05-23
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Hint: Let $G$ be a group $|G|=p^3$ and let $G$ act on this normal subgroup $N$ with $|N|=p^2$ by conjugation. This gives you a map $G\to \text{Aut}(N)$ whose kernel is the centralizer $C_G(N)$. Show that this map must be trivial so that $C_G(N)=G$ and so $N\leqslant Z(G)$. Note then that $G/N$ is cyclic to get a contradiction.

Hint: Let $G$ be a group $|G|=p^3$ and let $G$ act on this normal subgroup $N$ with $|N|=p^2$ by conjugation. This gives you a map $G\to \text{Aut}(N)$ whose kernel is the centralizer $C_G(N)$. Show that this map must be trivial so that $C_G(N)=G$ and so $N\leqslant Z(G)$. Note then that $G/N$ is cyclic to get a contradiction.

EDIT: As James Auld has pointed out this is clearly incorrect.

There is a theorem that says that the converse of Lagrange's theorem holds true for $p$-groups. So, if $G$ has order $p^3$ then it necessarily has a subgroup of order $p^2$. Since this subgroup has index the smallest prime dividing the order of the group it is definitely normal. Thus, the question is asking (secretly) if every group of order $p^3$ is abelian. This is definitely not true--think of non-trivial semi-direct products $\mathbb{Z}_p^2\rtimes_\varphi\mathbb{Z}_p$.

If you are interested in where my proof went wrong, it fell through because of a miscalculation of orders. The reason why I though the map $\phi:G\to\text{Aut}(N)$ was trivial was because $|G/\ker\phi|$ has to divide $(|G|,|\text{Aut}(N)|)$. Now, if we get $\mathbb{Z}_{p^2}$ then we have that $\text{Aut}(\mathbb{Z}_{p^2})\cong\mathbb{Z}_{p(p-1)}$. Clearly then this has order divisible by $p$ and so $(|G|,|\text{Aut}(N)|)$ has non-trivial divisors. Similarly, if we have $\mathbb{Z}_p^2$ then $\text{Aut}(N)\cong \text{GL}_2(\mathbb{F}_p)$ which has order $(p^2-1)(p^2-p)$ which is also divisible by $p$.

Either way, I was being crazy at computing orders. Hopefully my mistake has helped you learn though.