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Let $G$ be a group of order $12$. Show that if $Z(G)$ is non-trivial, then $G$ has a subgroup of order $6$.

By Cauchy's theorem, there are $a,b$ elements of $G$ of order $2,3$ and by action of conjugation, I know that $|G|=|Z(G)|+ \sum_{i=1}^{n}[G:C_{G}(g_i)]$, where $[g_i]$, $i=1,\dots,n$ are conjugacy classes with more than one element, but I can't find a way to show that there is a subgroup of order $6$.

Any suggestions?

citadel
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J P
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    Hint: If $H,K$ are subgroups and at least one of them is normal in $G$ then $HK$ is a subgroup as well. – Mark Nov 03 '23 at 13:30
  • Afterwards you can check directly here, because there are only two non-abelian groups with nontrivial center, namely $D_6$ and the dicyclic group of order $12$. Of course, both have a subgroup of order $6$. – Dietrich Burde Nov 03 '23 at 14:48

3 Answers3

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The center must contain an element $a$ of order $2$ or of order $3$.

Say $a$ has order $2$. Then $G$ has an element $b$ of order $3$. Because $a$ is central, $\langle a,b\rangle = \langle a\rangle\langle b\rangle$ has $6$ elements.

If $a$ has order $3$, then $G$ has an element $b$ of order $2$, and again we have that $\langle a,b\rangle = \langle a\rangle\langle b\rangle$ has order $6$.

citadel
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Arturo Magidin
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In $G$, there is an element $x$ of order $2$ and an element $y$ of order $3$. If $G$ is abelian, then $xy$ has order $6$.

If $|Z(G)|=6$, you are done.

If $|Z(G)|=4$, then necessarily $\langle x\rangle\le Z(G)$ (as otherwise $Z(G)\langle x\rangle$ would have order $8$, a contradiction); but then $xy$ has order $6$.

If $|Z(G)|=3$, then $Z(G)\langle x\rangle$ has order $6$.

If $|Z(G)|=2$, then $Z(G)\langle y\rangle$ has order $6$.

citadel
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    $Z(G)$ cannot have order $6$ (then $G/Z(G)$ would have order $2$). But it can have order $12$. – Arturo Magidin Nov 03 '23 at 20:29
  • Of corse, @ArturoMagidin, but one may not know that additional result. – citadel Nov 03 '23 at 21:32
  • And I have addressed the case $|Z(G)|=12$, so it's not quite clear what your last sentence refers to. – citadel Nov 03 '23 at 21:46
  • It reads like you are going through all the cases of $|Z(G)|$. But there is no explicit discussion of the case $|Z(G)|=12$ (it is only implicit in the first sentence, and easy to miss). – Arturo Magidin Nov 03 '23 at 21:57
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    You are missing the case where $|Z(G)|=4$. But actually, that case is impossible anyway, for the same reason that $|Z(G)|=6$ is impossible (namely, that $G/Z(G)$ is cyclic and so $G$ is abelian). – Geoffrey Trang Nov 03 '23 at 22:27
  • @GeoffreyTrang, you are right, Now I've edited, again in the spirit of not knowing the $G/Z(G)$ argument. – citadel Nov 03 '23 at 22:52
  • @citadel, more generally, if $G/Z(G)$ is cyclic, then $G$ is abelian, so if $|Z(G)|=4,6$, this would lead to $G/Z(G)$ being cyclic and then $G$ would be abelian, but if so, $|Z(G)|$=12, which is a contradiction. Is that so? – J P Nov 04 '23 at 10:54
  • Yes, @JP, it is so. – citadel Nov 04 '23 at 11:35
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For the nonabelian case you don't even need Cauchy, as the class equation suffices. If $Z(G)$ has order $2$, then$^\dagger$ the noncentral elements have centralizers of order $4$ or $6$, and there must be some of order $6$ as otherwise you couldn't "fill the gap" of $10(=12-2)$ with conjugacy classes of size $3(=12/4)$, only. If $Z(G)$ has order $3$, then$^\dagger$ the noncentral elements have centralizers of order $6$: as for your task, you'd be done, but note that all the noncentral conjugacy classes would have then size $2(=12/6)$, a contradiction because they cannot "fill the gap" of $9(=12-3)$; so, such a group doesn't exist. Finally, $Z(G)$ cannot have order $4$ or $6$, because$^\dagger$ $8\nmid 12$ and there's no divisor of $12$ strictly greater than $6$, respectively.

$^\dagger$For every noncentral element $x$, $Z(G)<C_G(x)<G$ (strictly).

citadel
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