Wreath Product of $\mathbf{Z}_{p^n}$ and $\mathbf{Z}_{p^m}$

Question

Is this true that: "The order of wreath product of $\mathbf{Z}_{p^n}$ and $\mathbf{Z}_{p^m}$ is $p^{m+n}$" ?

Answer 1

If $K$ acts on $\Omega$, then we can define an unrestricted wreath product of $G$ by $K$ (relative to $\Omega$) as follows: first, $G^{\Omega} = \{\mathbf{g}\colon\Omega\to G\}$ has the pointwise multiplication (it's the set of $\Omega$-tuples of elements of $G$). We define an action of $K$ on $G^{\Omega}$ by $$\mathbf{g}^k(\omega) = \mathbf{g}(k\cdot\omega).$$ This action gives a semidirect product $G^{\Omega}\rtimes K$; this semidirect product is called the (unrestricted) wreath product of $G$ by $K$, $G\wr_{\Omega}K$.

(The "restricted" wreath product is the subgroup in which $\mathbf{g}(\omega)=e$ for almost all $\omega\in\Omega$).

The order of $G^{\Omega}$ is $|G|^{|\Omega|}$, so the order of $G\wr_{\Omega}K$ is $|G|^{|\Omega|}|K|$.

When we don't specify $\Omega$ and there is no natural action of $K$, then we take $\Omega=K$ and let $K$ act on itself by left multiplication. Then we just write $G\wr K$, and the order of $G\wr K$ is then $|G|^{|K|}|K|$.

For $\mathbf{Z}_{p^n}\wr\mathbf{Z}_{p^m}$, the order is $|p^n|^{|p^m|}|p^m| = p^{m+np^m}$.

This is the same as the order of $(\mathbf{Z}_p)^n\wr(\mathbf{Z}_p)^m$, since the orders of the groups are the same, though the structure of the group is very different.

Answer 2

No. I assume you mean $Z_{p^m}$ as a permutation group on $p^m$ elements. In this case, the wreath product has order $(p^n)^{p^m}p^m$. In general, if $H$ is a permutation group on $n$ elements and $G$ is some group then the wreath product of $G$ and $H$ has order $|G|^n|H|$.