# Examples of infinite Semi-direct products

I'm looking for some examples of semi-direct products, $G = N \rtimes_\alpha H$ of (infinite) groups. I'm aware of the definitions involved but never really thought through a lot of examples.

I would mainly be interested in examples where $G$ is some kind of "easy" where $N$ and $H$ are "complicated".

To be more precise:

1) I am looking for examples where $H = BS(m,n)$ for $|m|,|n| \geq 2$ and $m \neq n$ ($H$ is the non-normal subgroup of $G$), i.e. $H$ is not residually finite.

2) I am looking for examples of an infinite group $G$ which "surprisingly" split as some semi-direct product.

Edit: I am only interested in finitely presented examples!

• $\mathbb{Z} \rtimes_{\phi} C_2 \cong H$ the infinite dihedral group, where $\phi(x)(n) = -n$? – Patrick Stevens Apr 8 '16 at 7:38
• May the downvoter please explain why he did downvote? Have I done something wrong in asking this question? – M.U. Apr 9 '16 at 6:33

## 3 Answers

I note that 1) seems to exclude the "solvable Baumslag-Solitar groups", which otherwise fit your question exactly. But I will describe them anyway, because they lead to a general class of examples.

Each such group $BS(1,n) = \langle a,t \, | \, tat^{-1}=a^n\rangle$ for $n \ge 2$ is isomorphic to the semidirect product $$\mathbb{Z}\bigl[\frac{1}{n}\bigr] \rtimes_\alpha \mathbb{Z}$$ where $\alpha(k) \in \text{Aut}(\mathbb{Z}[1/n])$ is the "multiplication by $n^k$" automorphism of the additive group structure on $\mathbb{Z}[1/n]$. Note that $\mathbb{Z}[1/n]$ is not even finitely generated as a group, which could be regarded as pretty complicated in the context of finitely presented groups.

More generally, any nonsurjective monomorphism $T : \mathbb{Z}^m \to \mathbb{Z}^m$ (i.e. any $n \times n$ integer matrix whose determinant has absolute value $\ge 2$) can be used to define a group similar to $BS(1,n)$, with $n$ commuting generators that together generate a $\mathbb{Z}^n$ subgroup, and with one more generator $t$ which conjugates each generator of the $\mathbb{Z}^n$ subgroup to its image under $T$. This group is a semidirect product of an infinitely generated rank $n$ abelian group with the infinite cyclic group.

• Nice, thank you. That's exactly the kind of example I wanted to see. – M.U. Apr 9 '16 at 6:23

(Example for (2)): Consider the group of (rigid) motions or isometries of Euclidean plane $\mathbb{R}^2$. The elements of this group are translations, rotations, reflections, and glide reflections. The translations form a normal subgroup, and there is another subgroup, called orthogonal subgroup, whose semi-direct product is the whole group. To be precise, any isometry of $\mathbb{R}^2$ is expressed by $$\begin{bmatrix} x \\ y\end{bmatrix} \mapsto A\begin{bmatrix} x \\ y\end{bmatrix} + \begin{bmatrix} a \\ b\end{bmatrix},$$ where $[x \,\, y]^t$ is an element of $\mathbb{R}^2$ and $A$ is $2\times 2$ orthogonal matrix.

Let $G$ deonte the group of all these maps under composition.

Let $N$ be the subgroup containing maps $$\begin{bmatrix} x \\ y\end{bmatrix} \mapsto \begin{bmatrix} x \\ y\end{bmatrix} + \begin{bmatrix} a \\ b\end{bmatrix}$$ and $H$ the subgroup containing maps $$\begin{bmatrix} x \\ y\end{bmatrix} \mapsto A\begin{bmatrix} x \\ y\end{bmatrix}.$$ Then $G=N\rtimes H$.

• Thanks for that! Please look at my edit, your answer made me refining the question accordingly. – M.U. Apr 8 '16 at 8:57

For (1), $G$ is necessarily not residually finite. This is because every subgroup of a residually finite group is residually finite. To see this, suppose $K\leq G$ and $G$ is residually finite, and let $k\in K$. Then there exists a homomorphism $\phi_k: G\rightarrow L_k$ where $L_k$ is finite. This induces a homomorphism $\phi_k|_K: K\rightarrow L_k$ and $k\not\in\ker(\phi_k)$ by construction.

For (2), I'll give you two examples which I find interesting.

Example 1. Suppose $G$ surjects onto $\mathbb{Z}$ with kernel $K$. Then $G\cong K\rtimes\mathbb{Z}$. (Note that $K$ is not necessarily finitely presented, or even finitely generated! For example, if $G$ is a (non-cyclic) for group then $K$ has infinite index in $G$, but all normal subgroups of infinite index in a free group are not finitely generated, by a relatively simple covering spaces argument.)

Example 2. It was an open problem for a very long time to classify when the cyclically presented groups \begin{align*} G_n&\cong\langle a_0, a_1, \ldots, a_n; a_0a_1=a_2, a_1a_2=a_3, \ldots, a_{n-2}a_{n-1}=a_0, a_{n-1}a_0=a_1\rangle\\ &=\langle a_0, a_1, \ldots, a_n; a_ia_{i+1\text{ mod }n}=a_{i+2\text{ mod }n}\rangle \end{align*} are finite. The original question was asked for $G_5$ as a "puzzle" by John Conway (it turns out $G_5$ is cyclic of order $11$, but this puzzle 3 years to solve!). Now, the cyclic group of order $n$, denoted $C_n$, acts on the generators in the obvious way so you can form the semidirect product $K_n=G_n\rtimes\mathbb{C_{n}}$. Then $K_n\cong \langle a, t; t^{-i}at^{-1}at^{-1}a^{-1}t^{i+2}, t^n\rangle$, which gives $$K_n\cong \langle a, t; a^2t^{-1}at^2, t^n\rangle.$$ If $n\geq11$ this is a $C(4)-T(4)$ "small cancellation" presentation (see the last chapter of Lyndon and Schupp's book Combinatorial Group Theory). This immediately implies that your semi-direct product $K_n$ is infinite. As $G_n$ has finite index in $K_n$ it must also be infinite. Hence, $G_n$ is infinite for all $n\geq 11$. (This proof is due to Roger Lyndon, and I found it (after much searching!) in the book "Presentations of Groups" by D.L.Johnson.)

• For (1): BS(m,n) with $m,n$ as in the question is not residually finite. So my $G$ is necessarily not residually finite. For your Example 1 for (2): Is there a way to generalize this by changing $\mathbb{Z}$ to some other group? – M.U. Apr 8 '16 at 11:18
• Sorry - that was a typo ( missing "not"). I proved what you've just said. – user1729 Apr 8 '16 at 11:22
• For your second point, I don't know off hand. $\mathbb{Z}$ works for two reasons - it has a copy $Z$ of itself in the preimage, and if $k\in K\cap Z$ is non-trivial then it "kills" part of $Z$ so $Z$ maps to a finite group, a contradiction. The second point is easy to mimick, e.g. a simple group or an almost finite group. I believe that the first point is the tough one to get working. – user1729 Apr 8 '16 at 11:28