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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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What have you tried? –  user21436 Apr 22 '12 at 2:48

4 Answers 4

up vote 1 down vote accepted

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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Sorry, but I cannot understand the following: 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. Can you clarify the matter? Thanks. –  awllower May 23 '12 at 2:48
    
@awllower: It is a standard result; I'm sure it's been asked here before, but: if $[K:H]=p$, then the action of $K$ on the left cosets of $H$ gives a map $K\to S_p$; if $p$ is the smallest prime that divides $|K|$, then the order of the image must be $1$ or $p$; it cannot be $1$ because the action is nontrivial, so the image has order $p$. Since the kernel is contained in $H$ and has the same index in $K$ as $H$, the kernel equals $H$ and therefore $H$ is normal (being the kernel of a homomorphism). –  Arturo Magidin May 23 '12 at 3:41
    
@awllower: here's a link to a proof in PlanetMath; I've been unsuccessful in locating it here. –  Arturo Magidin May 23 '12 at 3:46

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 edit –  Alex Youcis Apr 22 '12 at 3:10

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.

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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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Uh... Where does this useful result pop out? Thanks if any reference could be provided. –  awllower May 23 '12 at 2:50

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