Showing, for $G=\langle\delta\rangle\ltimes(A\times A)$ and abelian $A$, that $Z(G)=G'\cong A$.

This is the second part of Exercise 5.2.2(a) of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

Here is the previous part:

• Show $G=\langle\delta\rangle\ltimes D$ is nilpotent of class $2$.

The exercise is marked as being referred to later on in the textbook.

The Details:

(This can be skipped.)

On page 27 of Robinson's book,

[S]uppose that we are given two groups $H$ and $N$, together with a homomorphism $\alpha: H\to{\rm Aut}\, N$. The external semidirect product $G=H\ltimes_\alpha N$ (or $N\rtimes_\alpha H$) is the set of all pairs $(h,n), h\in H, n\in N$, with the group operation

$$(h_1,n_1)(h_2,n_2)=(h_1h_2,n_1^{h_2^\alpha}n_2).$$

The Question:

Let $A$ be a nontrivial abelian group and set $D=A\times A$. Define $\delta\in{\rm Aut}\, D$ as follows: $(a_1,a_2)^\delta=(a_1, a_1a_2)$. Let $G$ be the semidirect product $\langle\delta\rangle \ltimes D$.

(a) Prove that [. . .] $Z(G)=G'\cong A$.

(Here $Z(G)$ is the centre of $G$ (but Robinson uses the notation $\zeta(G)$) and $G'$ is the derived subgroup of $G$.)

Thoughts:

Since $A$ is abelian, so is $D$.

Notice that the first thing we need to show is an equality, not an isomorphism.

To this end, I could start with

$$Z(G)=\{ g\in G\mid gh=hg,\forall h\in G\}$$

and manipulate it to become

$$G'=\langle [x,y]\in G\mid x,y\in G\rangle,$$

where $[x,y]=x^{-1}y^{-1}xy$, or vice versa.

Another way is to let $g\in Z(G)$ then show $g\in G'$, then vice versa.

I'm not sure what to do.

Let $A=\Bbb Z_2$. Then $D$ is the Klein four group. Also ${\rm Aut}\, D\cong S_3$.

Now:

• $(0,0)^\delta =(0,0)$,

• $(0,1)^\delta=(0,1)$,

• $(1,0)^\delta=(1,1)$,

• $(1,1)^\delta=(1,0)$.

Thus $\delta^2$ is given by:

• $(0,0)^{\delta^2}=(0,0)$,

• $(0,1)^{\delta^2}=(0,1)$,

• $(1,0)^{\delta^2}=(1,0)$,

• $(1,1)^{\delta^2}=(1,1)$.

That is, $\langle \delta\rangle\cong \Bbb Z_2$.

Hence $G\cong \Bbb Z_2\ltimes (\Bbb Z_2\times\Bbb Z_2)$. Thus $G$ is either $D_4$ or $Q_8$, both of which have commutator and centre isomorphic to $\Bbb Z_2$.

The second thing we need to show is that $G'\cong A$.

I have no intuition here. If I had to guess what the centre is isomorphic to, I would have said $D$.

This is only a small part of the exercise. I think I should be able to answer it myself.

My reading of the book has been disrupted by my formal studies. I need a kick-start.

Please help :)

Solution 1:

Elements of the form $(e,a)|\,\, a\in A$ are central and form a group $K$ isomorphic to $A$. Also $$[\delta,(a,e)]=(a,a)(a,e)^{-1}=(e,a)$$

So we know $K\subseteq Z(G)$ and $K\subseteq G'$.

Also we know $K$ is a normal subgroup of $G$ and $$G/K\cong A\times \langle \delta \rangle$$ which is commutative, so $G'\subseteq K$. We can conclude $K=G'$.

If $\delta^i$ is a non-trivial automorphism of $A\times A$, then we have some $(x,y)\in A\times A$ with $$\delta^i(x,y)\neq (x,y).$$ Then for any $a,b\in A$ we have $(a,b)\delta^i$ does not commute with $(x,y)$.

Thus $Z(G)\subseteq A$ and the only elements of $A$ which commute with $\delta$ lie in $K$. Thus $Z(G)=K$.

Thus we have proven $Z(G)=G'\cong A$.