A non-abelian group of order 12 with only one element of order 2.

I require to show that, if I have a non - abelian team G of order 12 with just one component has order 2, after that G is soluble and also the facility Z (G) is such that

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

I do not recognize just how to address this trouble, and also any kind of aid would certainly be most welcome.

2022-07-25 20:43:52
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Answers: 1

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 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. $\Box$

2022-07-25 22:25:24