Proving every subgroup of a nilpotent group is subnormal

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!

• Don't we want $[HZ_i,HZ_{i+1}] \leq HZ_i$? – Mathematical Mushroom May 17 at 1:39
• 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. – Derek Holt May 17 at 7:20
• 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 – Mathematical Mushroom May 17 at 15:48

• Are you sure you are not assuming that $G$ is finite? – Derek Holt May 16 at 21:48