Understanding the wreath product of $\mathbb Z_2$ by $\mathbb Z_2$ and of $\mathbb Z_2$ by $\mathbb Z_3$

Question

1) What is a conceptual way to see that the regular wreath product of $\mathbb Z_2 \wr \mathbb Z_2$ is isomorphic to the dihedral group $D_4$ of order $8$?

How should I go about to answer this question? Should I first ask myself how to see that $D_4$ is a semidirect product? Or should I try to write down a presentation of $\mathbb Z_2 \wr \mathbb Z_2$?

1a) How do we see that the $2$-Sylow subgroup of the symmetric group $S_4$ is isomorphic to $\mathbb Z_2 \wr \mathbb Z_2$?

2) Now a general theorem says that one can find all extensions of a group $K$ by a group $H$ inside $K \wr H$. How do I in practice find subgroups of $D_4$ which are isomorphic to all extensions of $\mathbb Z_2$ by $\mathbb Z_2$? I mean in this example it is easy since there are two nonequivalent extensions, $\mathbb Z_2 \times \mathbb Z_2$ and $\mathbb Z_4$ and I look at the subgroup lattice of $D_4$ to find corresponding subgroups. But how would I go about this in general, e.g. in the next example.

3) Consider another example: $\mathbb Z_2 \wr \mathbb Z_3$. My understanding of this group is this: the base is $H = \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ with $G = \mathbb Z_3$ acting on the base by shifting the entries of a triple. Then the wreath product is the semidirect product $H \rtimes G$ wrt this shifting action. It is a group of order $2^3 \cdot 3 = 24$ elements. But this is about it how I understand this group. Are there easy observations to be made immediately about this group (just by definition)?

Answer 1

$\newcommand{\Set}[1]{\left\{ #1 \right\}}\DeclareMathOperator{\Sym}{Sym}$This is an answer to question (1)

What is a conceptual way to see that the regular wreath product of $\mathbb Z_2 \wr \mathbb Z_2$ is isomorphic to the dihedral group $D_4$ of order $8$?

Let $\Omega = \Omega_{1} \cup \Omega_{2}$ be a set which is the disjoint union of two sets $\Omega_{i}$ with $2$ elements each. Then $Z_{2} \wr Z_{2}$ is the group of all the permutations of $\Omega$ which respect the partition $\Set{\Omega_{1}, \Omega_{2}}$, that is, the set of all those $g \in \Sym(\Omega)$ that send $\Omega_{1}$ to either $\Omega_{1}$ or $\Omega_{2}$, and the same for $\Omega_{2}$.

Now $D_{4}$ is the group of congruences of a square. If the vertices of the square are labelled $1, 2, 3, 4$ consecutively (clockwise, say), then $D_{4}$ preserves the partition of the set $\Omega = \Set{ 1, 2, 3, 4 }$ of the vertices given by the two diagonals $\Omega_{1} = \Set{1, 3}$ and $\Omega_{2} = \Set{2, 4}$.

Now both groups have order $8$, so they must coincide.

Answer 2

At least for the $\mathbb{Z}_2 \wr \mathbb{Z}_2 \cong D_8$ case I can give you this proof:

We want to show that both groups have the same presentation and by that are isomorphic. For the $D_8$ we know the presentation from e.g. here. So $$ D_8 = \langle a, c \mid a^2, c^2, (ac)^2 \rangle $$ Also we will be using the fact, that a presentation of a direct product is given as $$ G \times H = \langle X \mid R \rangle \times \langle Y \mid S \rangle = \langle X \cup Y \mid R \cup S \cup T \rangle $$ with $T := \{ xyx^{-1}y^{-1} \} $

For the wreath product we also need a presentation of a semidirect product which can be found here (The proof can be found in the answer to this)

Now we consider the wreath product as the semidirect product of $S_2 \times S_2$ with $S_2$ as described i.e. here

So we get: $$ \begin{align} S_2 \wr S_2 &= (S_2 \times S_2) \rtimes S_2 \\ &= (\langle a, b \mid a^2, b^2, (ab)^2 \rangle ) \rtimes S_2 \end{align} $$

As we have $\phi:S_2 \to Aut(S_2 \times S_2), c^i \mapsto \left( (x, y) \mapsto \begin{cases} (x, y) &,i=1\\ (y, x) &,i=0,2 \end{cases} \right) $ where we consider the $S_2$ as cyclic.

Now we get by the semidirect product presentation definition $$ (S_2 \times S_2) \wr S_2 = \langle a,b,c \mid a^2, b^2, (ab)^2, c^2, cac=b, cbc=a \rangle $$ Now we eliminate the $b$ with $b=cac$: $$ \begin{align} (S_2 \times S_2) \wr S_2 &= \langle a, b,c \mid a^2, caccac, acacacac, c^2 caccac, ccacca, acac, ccacccac \rangle\\ &= \langle a,c \mid a^2, acacacac, c^2\rangle \\ &= \langle a,c \mid a^2, c^2, (ac)^4\rangle \end{align} $$

Now we introduce another parameter $d$ and relation $d=ac$ and get $$ \begin{align} &\langle a,c,d \mid a^2, c^2, (ac)^4, d=ac\rangle \\ =&\langle a,c,d \mid a^2, (ad)^2, d^4, d=ac\rangle \\ =&\langle a,d \mid a^2, (ad)^2, d^4\rangle \cong D_8 \end{align} $$

This can probably also be done for $\mathbb{Z_2} \wr \mathbb{Z}_3$ but I have not tried. If anybody wants to, I recommend GAP for checking your work, as it can do Tietze Transformations with comments.