I need to show that, if I have a non-abelian group G of order 12 with only one element has order 2, then G is soluble and the center Z(G) is such that

$Z(G)\cong \mathbb{Z}_2$ and $\frac{G}{Z(G)}\cong S_3$

I don't know how to solve this problem, and any help would be most welcome.

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    $\begingroup$ Doesn't one need $|G| = |Z(G)| |\frac{G}{Z(G)}|$? $\endgroup$ – Elchanan Solomon May 15 '12 at 14:13
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    $\begingroup$ Conjugation preserves order, so if you only have one element of order $2$ then it must be central. Now, there are only two groups of order $6$, $S_3$ and the cyclic group of order $6$. It is not too difficult to show that if $G/Z(G)$ is cyclic then $G$ is abelian. So, here you must have $G/Z(G)\cong S_3$. $\endgroup$ – user1729 May 15 '12 at 14:26
  • $\begingroup$ @DylanMoreland, you are correct, that was a typo. I have edited my text. $\endgroup$ – Marra May 15 '12 at 16:01
  • $\begingroup$ @DylanMoreland, but the order of $S_3$ is $3!$, so everything is fine $\endgroup$ – leo May 16 '12 at 3:34
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    $\begingroup$ Reposting: This KCd handout is useful for all of your groups of order 12 needs. $\endgroup$ – Dylan Moreland May 16 '12 at 3:53

Lemma 1. Let $G$ be a group, and let $N$ be a subgroup of $Z(G)$. If $G/N$ is cyclic, then $G$ is abelian.

Proof. Let $x\in G$ be such that $xN$ generates $G/Z(G)$. Let $a,b\in G$, and let $r,s\in\mathbb{Z}$ be such that $x^rN=aN$ and $x^sN=bN$. Then there exist $n,n'\in N$ such that $a=x^rn$ and $b=x^sn'$. We have: $$\begin{align*} ab &= (x^rn)(x^sn')\\ &= x^r(nx^s)n'\\ &= x^rx^snn' &\text{(since }N\subseteq Z(G)\text{)}\\ &= x^sx^rn'n &\text{(since }x^rx^s=x^sx^r\text{ and }n\in N\subseteq Z(G)\text{)}\\ &= x^sn'x^rn &\text{(since }n'\in N\subseteq Z(G)\text{)}\\ &= ba. \end{align*}$$ Thus, for all $a,b\in G$ we have $ab=ba$, so $G$ is abelian, as claimed. $\Box$

Lemma 2. Let $G$ be a group, and let $n$ be a positive integer. If $G$ has exactly one element of order $n$, then that element is central.

Proof. If $a$ has order $n$, then so does $xax^{-1}$ for all $x\in G$. Hence, since $a$ is the unique element of order $n$, we have $xax^{-1}=a$, or $xa=ax$. Thus, $a\in Z(G)$. $\Box$

(Added: As Jason de Vito points out in comments below, it's a bit of a silly lemma in that the only values of $n$ for which this can hold are $n=1$, in which case the conclusion is trivial, and $n=2$, which is the case at hand.)

Now, if $G$ has a unique element of order $2$, $a$, then $a$ is central by Lemma 2. Thus, $\langle a\rangle$ is normal, and $G/\langle a\rangle$ has order $6$; thus, $G/\langle a\rangle\cong S_3$ or $G/\langle a\rangle\cong \mathbb{Z}_6$. If $G/\langle a\rangle\cong\mathbb{Z}_6$, then by Lemma 1 we have that $G$ is abelian. Since $G$ is assumed to be nonabelian, we have $G/\langle a\rangle\cong S_3$.

Moreover, $Z(S_3)$ is trivial; since the image of the center of $G$ in $G/\langle a\rangle$ is central, it follows that $Z(G)=\langle a\rangle$. Thus, $Z(G)\cong\mathbb{Z}_2$ and $G/Z(G)\cong S_3$, as claimed.

To show that $G$ is solvable, we can now produce a normal series with abelian quotients. Let $g\in G$ be such that $gZ(G)$ generates the cyclic group of order $3$ in $S_3$. Then we have $$1\triangleleft \langle a\rangle\triangleleft \langle a,g\rangle \triangleleft G$$ where $\langle a,g\rangle\triangleleft G$ follows from the isomorphism theorems because $\langle gZ(G)\rangle\triangleleft G/Z(G)$. The quotients are: $$\begin{align*} \frac{G}{\langle a,g\rangle} &\cong \frac{G/\langle a\rangle}{\langle a,g\rangle/\langle a\rangle}\\ &\cong \frac{S_3}{\langle (123)\rangle}\\ &\cong\mathbb{Z}_2,\\ \frac{\langle a,g\rangle}{\langle a\rangle} &\cong \mathbb{Z}_3\\ \langle a\rangle &\cong \mathbb{Z}_2 \end{align*}$$ so all quotients are abelian, hence $G$ is solvable.

Alternatively, we know $S_3$ is solvable (since $[S_3,S_3]\cong\mathbb{Z}_3$ is abelian), and $\mathbb{Z}_2$ is solvable; and an extension of a solvable group by a solvable group is solvable. $\Box$

  • $\begingroup$ That is rather nice Arturo, although if you're going to go so far as to prove the lemmas, perhaps you ought to actually answer the question in full the asker posed, and show that G is solvable as well (my point being there's at least one more non-trivial lemma involved). $\endgroup$ – David Wheeler May 15 '12 at 18:40
  • $\begingroup$ @David: Fair enough! $\endgroup$ – Arturo Magidin May 15 '12 at 18:43
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    $\begingroup$ I just wanted to point out to the OP that "exactly one element of order $n$" is only possible when $n=2$ or $1$, for if $g$ has order $n$, so does $g^{-1}$. $\endgroup$ – Jason DeVito May 15 '12 at 18:54

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