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I know that a group of order 32 with $|Z(G)|=4$ has an abelian subgroup of order 16 using GAP (Groups, Algorithms, Programming). Is there a way without using programming to show this. Any help will be highly appreciated.

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  • $\begingroup$ This doesn't seem right. Are you sure this is always true? $\endgroup$ – Alex Youcis Sep 30 '13 at 7:57
  • $\begingroup$ @Alex Youcis. Yes it is true. $\endgroup$ – bor Sep 30 '13 at 7:59
  • $\begingroup$ I would imagine that the proof will use, at some point, the fact that if a group mod its center is cyclic, then it is abelian. $\endgroup$ – Alex Youcis Sep 30 '13 at 8:21
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    $\begingroup$ Hmm, so clearly we are done unless $G/Z(G)$ is elementary abelian. There is probably a good reason (possibly involving the usual Frattini subgroup arguments) for why this case also works out. $\endgroup$ – Tobias Kildetoft Sep 30 '13 at 8:35
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The result is clear if $G/Z(G)$ has an element of order 4, so assume $G/Z(G)$ is elementary abelian, and let $a,b,c$ be inverse images of its generators. I'll assume that $Z(G) = \langle x,y \rangle$ is elementary abelian. The case when it is cyclic is easier.

If $[a,b] = 1$, then $a,b,Z(G)$ generate an abelian subgroup of order 16. If $[a,b]=[b,c]$, then $[ac,b] = 1$ and we are done. So we can assume that $[a,b]=x$, $[a,c]=y$, $[b,c]=xy$. Then $[a,bc]=[b,bc]=xy$, so $[ab,bc]=1$.

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Note: This answer was made to a previous version of the question, which claimed that a group of order $32$ always has an abelian subgroup of order $16$.


This is not true and there are several counterexamples.

For example, you can verify that the GAP command $\operatorname{SmallGroup}(32, 6)$ returns a group of order $32$ with no abelian subgroup of order $16$.

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  • $\begingroup$ Indeed, you can construct one by hand I believe as $(\mathbb{Z}_4\rtimes\mathbb{Z}_4)\rtimes\mathbb{Z}_2$. $\endgroup$ – Alex Youcis Sep 30 '13 at 8:07
  • $\begingroup$ I am so sorry, i forgot an additional condition, that is $|Z(G)|=4$. Now i have added. $\endgroup$ – bor Sep 30 '13 at 8:13

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