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

Question

This is 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.

It 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

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

Robinson's definition of the following is equivalent to the one given on proof wiki:

Let $G$ be a group whose identity is $e$.

A normal series for $G$ is a sequence of normal subgroups of $G$:

$$\{e\}=G_0\lhd G_1\lhd\dots\lhd G_n=G,$$

where $G_{i-1}\lhd G_i$ denotes that $G_{i-1}$ is a proper normal subgroup of $G_i$.

On page 122 of Robinson's book,

Definition: A group $G$ is called nilpotent if it has a central series, that is, a normal series $1=G_0\le G_1\le \dots \le G_n=G$ such that $G_{i+1}/G_i$ is contained in the centre of $G/G_i$ for all $i$. The length of a shortest central series of $G$ is the nilpotent class of $G$.

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 $G$ is nilpotent of class $2$.

[NB: There is more to part (a) than I have shared here.]

Thoughts:

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

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

Now:

• $(0,0)^\delta =(0,0)$,

• $(0,1)^\delta=(0,0)$,${}^\dagger$

• $(1,0)^\delta=(1,0)$,

• $(1,1)^\delta=(1,1)$.

I can't see the wood for the trees.

Finding $\langle\delta\rangle$ seems intractable, at least using the naïve calculations above.

Since $\delta$ is nontrivial and ${\rm Aut}\, D\cong \Bbb Z_3\rtimes\Bbb Z_2$ is not cyclic, we have either $\langle \delta\rangle\cong\Bbb Z_3$ or $\langle \delta\rangle\cong\Bbb Z_2$.

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I'm at a loss as to how to do this for arbitrary $A$.

Perhaps calculating $\langle\delta\rangle$ is unnecessary . . . I'm not sure.

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This question is only a small part of the exercise, so it has me thinking I ought to be able to answer it.

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Please help :)

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$\dagger$ I have just realised a couple of weeks later that these calculations are wrong, hence the trouble.

Answer 1

Note that saying that the group $G$ is nilpotent $G$ is equivalent to saying that the sequence of subgroups defined by $$\gamma_{1}(G) = G, \qquad \gamma_{i+1}(G) = [\gamma_{i}(G), G] \text{, for $i \ge 1$}$$ terminates to $\{1\}$. Here $[H, K] = \langle [h, k] : h \in H, k \in K \rangle$, where $[h, k] = h^{-1} k^{-1} h k$ is the commutator of $h$ and $k$.

Note that the commutator subgroup $\gamma_{2}(G) = [G, G]$ of $G$ is contained in $D$, as $G/D$ is cyclic.

Now consider the commutator $$ [(a_{1}, a_{2}), \delta] = (a_{1}, a_{2})^{-1} (a_{1}, a_{2})^{\delta} = (a_{1}^{-1}, a_{2}^{-1}) (a_{1}, a_{1} a_{2}) = (1, a_{1}). $$ Therefore $$ [(a_{1}, a_{2}), \delta, \delta] = 1. $$ Please let me know if this keeps you going.