This is Exercise 1.6.14 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE.
The Details:
The definition of the wreath product given in the book is quite intricate.
On page 32 and 33, ibid.,
Let $H$ and $K$ be permutation groups on sets $X$ and $Y$ respectively. [. . .]
If $\gamma\in H$, $y\in Y$, and $\kappa\in K$, define
$$\gamma(y):\begin{cases} (x,y) \mapsto (x\gamma, y) & \\ (x,y_1) \mapsto (x,y_1) & y_1\neq y,\end{cases}$$
and$$\kappa^\ast: (x,y)\mapsto (x,y\kappa).$$
[. . .] The functions $\gamma\mapsto \gamma(y)$, with $y$ fixed, and $\kappa\mapsto \kappa^\ast$ are monomorphisms from $H$ and $K$ to ${\rm Sym}(X\times Y)$: let their images be $H(y)$ and $K^\ast$, respectively. Then the wreath product of $H$ and $K$ is the permutation group on $X\times Y$ [. . .]
$$H\wr K=\langle H(y), K^\ast\mid y\in Y\rangle.$$
[. . .]
[T]he base group $B$ of the wreath product [is]
$$B=\underset{y\in Y}{{\rm Dr}}\, H(y).$$
Here $\underset{y\in Y}{{\rm Dr}}\, H(y)$ is defined as in this question.
Standard Wreath Products:
On page 41 of the book,
If $H$ and $K$ are arbitrary groups, we can think of them as permutation groups on their underlying sets via the right regular representation and form their wreath product $W=H\wr K$: this is called the standard wreath product.
The Question:
Let $W=H\wr K$ be the standard wreath product of an abelian group $H\neq 1$ and an arbitrary group $K$. Prove that the centre of $W$ equals the set of elements in the base group all of whose components are equal. (This is called the diagonal subgroup of the base group.)
Thoughts:
I'm confused.
I'll work through an example.
Let $H=\Bbb Z_3$ and $K=D_3$, the dihedral group of order $6$. Then $W=H\wr K$ is given by viewing $H$ and $K$ acting on their underlying sets, like so:
$$h_H^\rho: x\mapsto xh\text{ and }k_K^\rho: x\mapsto xk.$$
Let $y\in \Bbb Z_3$ be fixed. I need to specify $\gamma(y)$ and $\kappa^\ast$, for $\gamma\in \Bbb Z_3$ and $\kappa\in D_3$, as in the definition above.
What do I do next?
The diagonal subgroup for $\Bbb Z_3\wr D_3$ is unclear to me, let alone for $H,K$ as in the question.
I chose $\Bbb Z_3$ because it is cyclic and $D_3$ because it is the smallest nonabelian group. These might be bad choices, seeing as though the standard wreath product, according to GroupNames, has order $162$.
Please help 🙂
Best Answer
I’m going to use the following description of $H\wr K$: the “base group” is $H^K$, the direct product of $|K|$ copies of $H$, indexed by $K$. An element will “look like” $(h_k)_{k\in K}$. It has its usual group structure.
We let $K$ act on $H^K$ as follows: given $r\in K$, $(h_k)_{k\in K}\in H^K$, we let $(h_k)^r = (h_{kr})_{k\in K}$. That is, $K$ is acting on the indices of $H^K$.
Then $H\wr K = H^K\rtimes K$, using this action.
So the elements look like pairs $((h_k)_{k\in K}, r)$ with $h_k\in H$, $r\in K$, and product $$\bigl( (h_k)_{k\in K},r\bigr)\bigl( (g_k)_{k\in K},s\bigr) = \bigl( (h_k)_{k\in K}((g_k)_{k\in K})^{r^{-1}}, rs\bigr) = \bigl( (h_kg_{kr})_{k\in K},rs\Bigr).$$
I’m sure it won’t be hard to translate this description into yours (it’s possible one may need to switch the action from a left action to a right action, or change the shuffling as being done by $r^{-1}$ instead of $r$; sorry, but I don’t have the mental bandwidth right now to try to reconcile my usual picture of the wreath product with the description you are giving...)
First, we prove that if $((h_k)_{k\in K},r)$ is central, then $r=e$. Indeed, let $h\neq e_H$, and consider the element $\mathbf{h}\in H^K$ which has $h$ in the $e_K$ coordinate, and $e_H$ elsewhere. Then $$(\mathbf{h},e_K)\bigl((h_k)_{k\in K},r\bigr)$$ has second coordinate $r$, and first coordinate equal to $h_k$ in the coordinates $k\in K$, $K\neq e$, and equal to $hh_e$ in the $e$ coordinate (because $e_K$ acts trivially on $(h_k)_{k\in K}$).
On the other hand, $$\bigl( (h_k)_{k\in K},r\bigr)(\mathbf{h},e_K)$$ has first coordinate that has $h_{r^{-1}}h$ in the $r^{-1}$th coordinate, and $h_k$ in all $k$th coordinates with $k\neq r^{-1}$ (because $r^{-1}$ acted on $\mathbf{h}$ by moving $h$ to the $r^{-1}$ coordinate, and all other coordinates are trivial). Since we assumed that $((h_k),r)$ was central, the only way these two products are equal is if $r=e$. Thus, all central elements are in the “base” group, $H^K$.
Now let $((h_k),e)$ be an element of the base. Assume that there exist $k_1\neq k_2$ in $K$ with $h_{k_1}\neq h_{k_2}$. Then the product $$((e),k_2^{-1}k_1)((h_k),e)$$ will have a first coordinate which is a tuple with $h_{k_1}$ in the $k_2$ coordinate; whereas $$( (h_k),e) ((e),k_2^{-1}k_1)$$ will have $h_{k_2}$ in the $k_2$-coordinate of the base element. Thus, this element is not central.
So the only possible central elements are those of the form $((h)_{k\in K},e)$, where $(h)_{k\in K}$ is the tuple that has $h$ in all coordinates. Then it is a matter of verifying that they are central. But indeed, we have $$( (h)_{k\in K},e)\bigl( (h_k),r\bigr) = \bigl( (hh_k)_{k\in K},r\bigr)$$ and $$\bigl( (h_k)_{k\in K},e\bigr)( (h)_{k\in K},e) = \bigl( (h_kh)_{k\in K},r\bigr)$$ (because the “constant tuple” doesn’t care about the reshuffling), and because $H$ is assumed to be abelian, these two elements are equal.
(You can probably guess from this that for an arbitrary $H$, the center will instead consist of the elements with trivial $K$ coordinate, and where the base element is a constant tuple with entry in $Z(H)$ )