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

Question

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?

Answer 1

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.

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

Answer 2

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 $6$.

Answer 3

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 $6$.