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Let $G$ be a $p$-group contained in $S_{p^m}$ for some $m\in \mathbb{N}$, this is, the symmetric group of degree $p^m$. Assume that the lower central series, LCS for short, of $G$ is well understood.

Is there anything known about the LCS of $W(G):=G\wr C_p$, where $C_p$ is a cyclic group of order $p$, and $\wr$ is the (right regular) permutation action of $C_p$ on $G\times \dotsb\times G$. That is, if $(g_1,g_2,\dotsc, g_p)\in G\times \dotsb\times G$, then the action of $C_p=\langle x\rangle$ on the base group $G\times \dots\times G$ is defined by $(g_1,g_2,\dotsc, g_p)^x=(g_p,g_1,\dotsc,g_{p-1})$.

I am particularly interested in the indices $|\gamma_i(W(G)):\gamma_{i+1}(W(G))|$ for $i \in \mathbb{N}$.

I have been searching for some bibliography or papers on the topic but I have not been able to find anything.

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    $\begingroup$ The case in which $G$ is an abelian $p$-group is already not quite trivial, and it was studied implicitly by Hall in the last section of doi.org/10.1017/S0305004100031662. In this case the indices are $|G|p, |G|, |G|, \dots, |G|$. If $G$ is not a $p$-group then $W(G)$ need not be nilpotent, e.g., $C_3 \wr C_2 \cong C_3 \times S_3$ (but your question is still reasonable). $\endgroup$ Mar 6, 2023 at 15:25
  • $\begingroup$ Thank you very much! But in my case, $G$ is a $p$-group and $C_p$ to. What I mean is that both of the groups are powers of the same $p$, so the case $C_3\wr C_2$ is not considered. In fact it can be seen that $W(G)$ is always nilpotent, in my case. $\endgroup$
    – Gillyweeds
    Mar 6, 2023 at 15:47
  • $\begingroup$ Sorry, I missed the condition that $G$ is a $p$-group. $\endgroup$ Mar 7, 2023 at 9:36
  • $\begingroup$ I think my previous comment is only correct if $G$ actually elementary abelian. $\endgroup$ Mar 7, 2023 at 15:43
  • $\begingroup$ Since you don't seem to use any particular embedding $G \to S_{p^m}$, it's not necessary to state it, right?—since the bare existence of such an embedding follows from Cayley's theorem, with $p^m = \lvert G\rvert$. $\endgroup$
    – LSpice
    Mar 10, 2023 at 19:51

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Here is a solution in the special case in which each lower central factor $\gamma_i(G) / \gamma_{i+1}(G)$ is elementary abelian.

First consider the case in which $G$ is elementary abelian, written additively, so $G \cong \mathbf F_p^d$ for some $d$. Let $x$ be a generator of $C_p$. Then for $g \in G^p$ we have $[g,x] = \Delta(g)$, where $\Delta = \sigma-1$ and $\sigma : G^p \to G^p$ is the linear map defined by $\sigma(g_1, \dots, g_p) = (g_p, g_1, \dots, g_{p-1})$. Therefore $$\gamma_i(G \wr C_p) = \Delta^i(G^p) \qquad (i \ge 2).$$ By the $\mathbf{F}_p[X]$ polynomial identity $(X-1)^{p-1} = 1 + X + \cdots + X^{p-1}$ we have $$\Delta^{p-1} = 1 + \sigma + \cdots + \sigma^{p-1}.$$ Therefore $$\Delta^{p-1}(g) = (s, \dots, s), \qquad \text{where}~s=g_1 + \cdots + g_p.$$ Therefore $\gamma_{p-1}(G \wr C_p) = \Delta^{p-1}(G^p)$ is just the diagonal copy of $G$ in $G^p$, which is also the kernel of $\Delta$. It follows that the lower central factors of $G \wr C_p$ are $G \times C_p, G, \dots, G$.

Now consider the general case. Let $G = \Gamma_1 \ge \cdots \ge \Gamma_k = 1$ be the lower central series of $G$. Then $C_p$ acts on each factor $\Gamma_i^p / \Gamma_{i+1}^p$, and the lower central series of $\Gamma_i / \Gamma_{i+1} \wr C_p$ is as described above. By lifting to $G$ we get a long central series: $$G \wr C_p = \Gamma_1^p C_p \ge \Delta(\Gamma_1^p) \Gamma_2^p \ge \Delta^2(\Gamma_1^p) \Gamma_2^p \ge \cdots \ge \Delta^{p-1}(\Gamma_1^p) \Gamma_2^p \ge \Gamma_2^p \ge \Delta(\Gamma_2^p) \Gamma_3^p \ge \cdots.$$ I claim this is in fact the lower central series. To see this it suffices to observe that $\Delta^i(\Gamma_j^p)\Gamma_{j+1}^p$ (for $i \le p-1$) contains the preimage $D$ in $\Gamma_j^p$ of the diagonal subgroup of $\Gamma_j^p/\Gamma_{j+1}^p$, and $[D, G^p] = \Gamma_{j+1}^p$.

In general if $\gamma_i(G)/\gamma_{i+1}(G)$ is not elementary abelian it seems more complicated. For example $C_{p^2} \wr C_p$ seems to have lower central factors $C_{p^2} \times C_p, C_p, C_p, \dots, C_p$.

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  • $\begingroup$ I have a small question. At the beginning where do you use that $G$ is elementary abelian? Is it to get that the map $\Delta$ has $\dim \ker\Delta= \dim G$? $\endgroup$
    – Gillyweeds
    Mar 10, 2023 at 14:03
  • $\begingroup$ @Gillyweeds The polynomial identity $\Delta^{p-1} = 1 + \sigma + \cdots + \sigma^{p-1}$ uses the fact that $G$ has exponent $p$. $\endgroup$ Mar 10, 2023 at 15:27

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