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Mar 11, 2023 at 11:43 comment added Gillyweeds Yes, indeed, it is not required.
Mar 10, 2023 at 19:51 comment added LSpice 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 $p^m = \lvert G\rvert$.
Mar 10, 2023 at 19:50 history edited LSpice CC BY-SA 4.0
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Mar 8, 2023 at 18:18 vote accept Gillyweeds
Mar 7, 2023 at 16:24 answer added Sean Eberhard timeline score: 2
Mar 7, 2023 at 15:43 comment added Sean Eberhard I think my previous comment is only correct if $G$ actually elementary abelian.
Mar 7, 2023 at 9:36 comment added Sean Eberhard Sorry, I missed the condition that $G$ is a $p$-group.
Mar 6, 2023 at 15:47 comment added Gillyweeds 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.
Mar 6, 2023 at 15:25 comment added Sean Eberhard 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).
S Mar 6, 2023 at 14:19 review First questions
Mar 6, 2023 at 14:28
S Mar 6, 2023 at 14:19 history asked Gillyweeds CC BY-SA 4.0