Let say I have an infinite elementary abelian $p$-group $E$ (i.e. with presentation $E= \langle x_1,x_2,x_3,... \mid x_i^p=1, \ x_i x_j = x_j x_i \rangle$).

How do I find the Fitting subgroup of the wreath product $G := \mathbb{Z}_p \wr E$, or more precisely, how could I prove that $\operatorname{Fitt}(G) = G$?

What my thinking is, find a series of normal subgroups with increasing nilpotency class, so something like $\mathbb{Z}_p \times E$, $(\mathbb{Z}_p \oplus \mathbb{Z}_p) \times E, \ldots$ but these don't seem to be normal in $G$ and I can't think of any better ones.

Any ideas?


1 Answer 1


Let $B$ be the base group of the Wreath product. So $B$ is also an infinite elementary abelian $p$-group. Also, or $n \ge 0$, let $E_n$ be the finite subgroup $\langle x_1,\ldots,x_n \rangle$ of $E$.

Now the subgroups $H_n := \langle B, E_n \rangle$ for $n \ge 0$ are all normal in $G$. Note that $H_n$ is a subdirect product of infinitely many copies of $C_p \wr E_n$, and hence is nilpotent.

So since $G = \cup_{i \ge 0} H_n$, we have ${\rm Fit}(G) = G$.


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