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Let $G$ be a group of order $pq^2$, where $p \neq q$ prime and $p$ does not divide $| Aut (G) |$. Show that $G$ is abelian.

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If this is homework, have you at least tried something? Can you show some work? Note that this is proved in some algebra books exercises ; I know that in Dummit & Foote's Abstract Algebra, there's a sketch of proof in the exercises. –  Patrick Da Silva Jun 20 '12 at 4:40
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It would be nice if you didn't phrase your question as a command, if you told us what you know about the problem, what theorems you are aware of that might be relevant, what work you have put into it yourself, etc., etc., etc. –  Gerry Myerson Jun 20 '12 at 4:40
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Isn't it important to know that $p$>$q$ or $q$>$p$ in this problem? –  B. S. Jun 20 '12 at 6:23
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2 Answers

up vote 1 down vote accepted

Firstly,we can consider a homomorphism $f:G\rightarrow Aut(G)$ such that: $f(x)=t_x$,where $t_x:G\rightarrow G$ is defined by $t_x(g)=xgx^{-1}$.

Note that $ker f=Z(G)$,then by first isomorphism theorm, we get: $G/Z(G)\cong f(G) \le Aut(G)$.So $|G| $can be divided by $|Z(G)||Aut(G)|$.

Hence, $Z(G)$ is divided by $p$.By Cauchy Theorem,$Z(G)$ contains a element of order $p$. We find that the cyclic subgroup generated by that element is of order $p$ and hence is Sylow $p$ subgroup and since it is in the center, we can conclude that it is the unique Sylow $p$ subgroup,which is also normal.

Let $Q$ be a Sylow $q$ subgroup in $G$.Since $p$ and $q$ are relative prime,we have $P\cap Q=\{1\}$.Besides, $PQ$ is a subgroup of order $pq^2$ since $P$ is normal and $|PQ|=|P||Q|/|P\cap Q|$.We have $PQ=G$.

Now, we only focus on $Q$.From class equation ,we get that $Z(Q)$ is nontrivial.If $Z(Q)$ is of order $q^2$,then $Q$ commutes with its elements, otherwise $|Z(Q)|=q$. In this case,$Q/Z(Q)$ which is of order $q$ and hence is cyclic, so $Q$ is abelian. In both case, $Q$ always commutes with its elements.

Since $G=PQ$,$P$ is contained in the centre and $Q$ commutes with its elements,we can conclude that $G$ is abelian.

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I don't have that reference noted by Patrick, but I think you can use the following hint.

Hint: Use the fact that $|\frac{G}{Z(G)}|$ divides the order of $Aut(G)$.

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So $|G/Z(G)|$ is not divisible by p which implies $|Z(G)|$ is $p, pq$ or $pq^2$, the case $Z(G)=pq$ is ruled out as otherwise $G/Z(G)$ is cyclic and thus $Z(G)=G$. It remains to rule out the case $Z(G)=p$ ? –  user31899 Jun 20 '12 at 8:29
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@user31899: If $Z(G)=p$ then $G≅Z(G)×\mathbb Z_{q^2}$? –  B. S. Jun 20 '12 at 8:39
    
Good work, nice hint! +1 –  amWhy Mar 6 '13 at 0:58
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