I encountered the problem below:
Let $D\triangleleft A\wr B$. Prove that $D\cap K\neq 1$ if $D\neq 1$, where $K$ is the base group of $A\wr B$.
Firstly I cannot understand the meaning of $D\cap K$ here. So the (regular) wreath product $A\wr B=K\rtimes B=\{(k, \ b) \ | \ k\in K, \ b\in B\}$ with the multiplicaton defined as $$(k_1, \ b_1)(k_2, \ b_2)=(k_1k_2^{{b_1}^{-1}},\ b_1b_2).$$ And $D$ is a normal subgroup of the group above. So what could the set $D\cap K$ be? Can I just see $K$ as $K\times \{1\}$ where $1$ is the unit in $B$?
So suppose I am correct. Here we regard $K$ as $K\times \{1\}$ where ${1}$ is the multiplicative unit in $B$. Suppose there is some $1\neq(k,\ b)\in D$. Since $B$ is a finite group, $b\in B$ must have a finite order, say $m$. Then $(k,\ b)^m\in K\times \{1\}$. So $(k,\ b)^m\in D\cap K$. Now I am left to show that $(k,\ b)^m\neq 1$ holds for some $k$. And this turns out to be equivalent to $(kb)^m\neq 1$. I am stuck here in applying the fact that $D$ is normal, and I do not know whether this is the right way.
Thank you for your help!