If $|G|=12$ with $|Z(G)|=2$, prove that the group $G$ has only one subgroup of order $3$.

Question

I have to solve the following exercise:

If $G$ is a group of order $12$ with the center $Z(G)$ of order $2$, prove that $G$ has only one subgroup of order $3$.

My idea is, since $n_3$, the number of $3$-Sylow can be $1$ or $4$, if $n_3=4$, we have $8$ elements of order $3$ thus an unique subgroup of order $4$, so normal. For each element $x \notin Z(G)$, $|C(x)|=4$ or $|C(x)|=6$, where $|C(x)|=\{g \in G \;|\; gx=xg\}$. From the classes equation $$|G| = |Z(G)| + \sum_{x \notin Z(G)} \frac{|G|}{|C(x)|} \quad \Longrightarrow \quad 10 =3\lambda + 2\mu$$ with solution $(0,5)$ or $(2,2)$. So we have $5$ or $4$ different conjugacy classes. If I'm not wrong, all $3$-Sylow are conjugate thus all $8$ elements of order $3$ are in the same conjugacy class, the other two element out the center are in at most $2$ conjugacy classes. From this can I deduce the absurd?

There are just $3$ non-abelian groups of order $12$, so it is very easy to check this directly - see here. — Dietrich Burde, Dec 17 '21 at 16:41
I am interested in whether my reasoning is correct or if there is an easier way to get to the solution, without using examples. — Rick88, Dec 17 '21 at 16:43
Hint: if the centre has order $2$, then prove that there is an element of order $6$. Where does this fit if $n_{3} = 4$? — Andreas Caranti, Dec 17 '21 at 16:53
if $z \neq e$ is the unique element of the center and $n$ is an element of order $3$, is their product $zn$ an element of order $6$? — Rick88, Dec 17 '21 at 16:56
@DietrichBurde, yes but you would help me more if you find some error in my reasoning. — Rick88, Dec 17 '21 at 16:58
"All p-Sylow groups are conjugate" means if A and B are p-sylow subgroups then $A=gBg^{-1}$ for some $g\in G$. It does not follow that all elements in A and B are contained in the same conjugacy class — user950314, Dec 17 '21 at 17:10
@AndreasCaranti, I don't understand how use your hint, can you explain me the solution? — Rick88, Dec 17 '21 at 17:17
If $G/Z(G)$ is cyclic, then $G$ is abelian. So, $G/Z(G)$ must be isomorphic to $S_3$, the non-cyclic (and non-abelian) group of order $6$. — Geoffrey Trang, Dec 17 '21 at 17:21

score 1 · Answer 1 · answered Dec 17 '21 at 20:59

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Simpler argument: What is the order of normalizer of a sylow 3-group $H$? It contains $H$ and $Z(G),$ so has to be divisible by $6.$ So, it's index is $1$ or $2$. Since we cannot have $2$ Sylow 3-subroups, we must have $1.$

answered Dec 17 '21 at 20:59

Igor Rivin

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score 1 · Answer 2 · answered Aug 30 '23 at 21:05

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By contradiction, suppose there are two of them, say $H_1$ and $H_2$. Then, $K_1:=Z(G)H_1$ and $K_2:=Z(G)H_2$ are two subgroups of order $6$ which intersect in $Z(G)$, whence: $|K_1K_2|=6.6/2=18$: contradiction, because $K_1K_2\subseteq G$.

answered Aug 30 '23 at 21:05

citadel

2,940

score 0 · Answer 3 · answered Dec 17 '21 at 20:35

$G$ be a group of order $12$ with $Z(G) =2$.

$|G|=12=2^2•3$

$P:2-SSG $

$Q:3-SSG$

Then, \begin{align}n_3 &=(1+3k) |2^2 \space , k=0, 1,...\\ &=1 \text{ or } 3 \end{align}

Claim: $n_3 =1$

Suppose, $n_3 =4$, then by Sylow's second theorem $G $ acts transitively by conjugation on the set of sylow $3$-subgroups. This action gives us a group homomorphism, $\phi : G \to S_4$

$Q\lhd N_G(Q)$ and and $[G:N_G(Q) ]=n_3$

$\implies N_G(Q) =Q$

Now, $Ker(\phi) \subseteq Stab_G(Q) $ is a normal subgroup of $G$.

Here, $Stab_G(Q) =N_G(Q) $ because the group action of $G$ on the set of $Q's$ is conjugation.

Since$|Q|=3 \implies Ker(\phi) =\{e_G\}$

Hence, $\phi:G\to S_4$ is an one to one homomorphism.

And, $[S_4 :G]=2 \implies G\lhd S_4$

Hence, $G=A_4$

Then $|Z(G)| =|Z(A_4)| = 1$ , a contradiction.

Hence, $n_3 =1 \implies P$ is unique.

If $|G|=12$ with $|Z(G)|=2$, prove that the group $G$ has only one subgroup of order $3$.

3 Answers3