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Let $G$ be a nilpotent group, i.e. $\exists$ an upper central series:

$1\triangleleft Z_1(G)\triangleleft Z_2(G)\triangleleft ....\triangleleft Z_k(G)=G$

Where $\frac{Z_{i+1}}{Z_i}=Z(\frac{G}{Z_i})$

To show that every subgroup of $G$ is subnormal, let $H \leq G$. Then we can multiply every term of the upper central series of $G$ by $H$ to get a new series:

$H\leq HZ_1(G)\leq HZ_2(G)\leq ....\leq HZ_k(G)=G$

If we can show that $HZ_i(G)\triangleleft HZ_{i+1}(G)$ then we are done.

I am having trouble doing this. There was another post on stack exchange similiar to this from awhile back but I found the answers inadequate so I thought i'd ask again. Thanks!

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  • $\begingroup$ Don't we want $[HZ_i,HZ_{i+1}] \leq HZ_i$? $\endgroup$ – Mathematical Mushroom May 17 at 1:39
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    $\begingroup$ You seem to have written $\rhd$ where you mean $\lhd$. I think you mean $1 \lhd Z_1(G) \lhd Z_2(G)$ and you want to prove $HZ_i \unlhd HZ_{i+1}$. So $[HZ_i,HZ_{i+1}] = [H,H][H,Z_i][H,Z_{i+1}][Z_i,Z_{i+1}] \le HZ_i$ as required. $\endgroup$ – Derek Holt May 17 at 7:20
  • $\begingroup$ Thank you. You are correct, it was my mistake, I am going to edit my question, so if you want to edit your response go for it. That is a great way to show normality, by far the simplest I've seen $\endgroup$ – Mathematical Mushroom May 17 at 15:48
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Hint (in case of finite groups): use the “normalizers grow” principle in nilpotent groups.

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    $\begingroup$ Are you sure you are not assuming that $G$ is finite? $\endgroup$ – Derek Holt May 16 at 21:48
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    $\begingroup$ Yes, my bad, I see there is no finiteness condition. Will adjust. $\endgroup$ – Nicky Hekster May 17 at 5:20

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