## Quotient of subgroups by center.

Let $H \leq G$. Let $Z_G$ denote the center $[G,G]$ the commutator subgroup. Assume $[G,G] \leq Z_G$ (i.e. nilpotent of class 2). Then $G/Z_G$ is abelian since $Z_G$ contains the commutator subgroup. There is a theorem that states that $H$ must also be nilpotent of class at most 2, hence $[H,H] \leq Z_H$ so $H/Z_H$ is also abelian.

A few questions:

1) Suppose $G/Z_G$ $\simeq$ $\mathbb{Z}^k$ and $H/Z_H$ $\simeq$ $\mathbb{Z}^t$

Is it always the case that $t \leq k$?

2) What if $Z_G$ and $Z_H$ are replaced by $[G,G]$ and $[H,H]$ respectively?

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 No. I am trying to analyze the graded Lie ring G/Z + Z_G by interpreting subgroups of G as subrings of the form H/Z_H + Z_H. In other words, I am trying to find a useful map into the integers. So I want to know if the ranks stay 'split' or if there is some overlap. I think there is a problem already, though, since I think it is not necessarily the case that Z_H < Z_G – Stuart May 19 2011 at 17:02 @Stuart: It is indeed not necessarily the case that $Z_H\lt Z_G$; for a trivial example, take $H=\langle h\rangle$, with $h\notin Z_G$. – Arturo Magidin May 19 2011 at 17:11 The main thing is obviously when you have a 'very abelian' subgroup in a 'very non-abelian' group or vice versa. But I can't tell whether the structure (e.g. nilpotent of class two) is enough to save me. – Stuart May 19 2011 at 17:11 The important point is that $Z_H \geq Z_G \cap H$, so you have canonical maps from $H/(Z_G \cap H)$ to both sides. One is an injection, and the other is a surjection. – S. Carnahan♦ May 19 2011 at 17:19 @Arturo. Sorry, what I meant to say was whether the rank of the isomorphic image of $Z_H$ is smaller than the rank of the isomorphic image of $Z_G$. And you may be right. The thing is, since it is nilpotent of class two, it ends up being that the commutator subgroups lie inside the centers. Since the commutator subgroup of H is just bashing out all possible commutators of elements of H rather than of G, it must be that [H,H] $\leq$ [G,G]. So it might also happen to be the case for the centers. On the other hand, there is the option of instead using G/G' + G' and H/H' + H', as in question 2. – Stuart May 19 2011 at 17:20

The answer to (1) is "yes." As Carnahan notes, you have $H\cap Z_G\subseteq Z_H$. Since $H/(Z_G\cap H)\cong HZ_G/Z_G$ is isomorphic to a subgroup of $G/Z_G$, then, $H/Z_H$ is a quotient of $H/(Z_G\cap H)$, hence isomorphic to a quotient of a subgroup of $G/Z_G$, so $\mathrm{rank}(H/Z_H) \leq \mathrm{rank}(G/Z_G)$.

The answer to (2) is "no". Take $G$ to be the relatively free group of class $2$ and rank $k$ (isomorphic to $F_k/(F_k)_3$, where $F_k$ is the free group of rank $k$ and $(F_k)_3$ is the third term of the lower central series of $F_k$). Then $G^{\rm ab}\cong \mathbb{Z}^k$. Let $H=[G,G]$; then $H$ is free abelian of rank $\binom{k}{2}$. Pick $k\gt 3$ to get that the rank of $H/[H,H]$ is greater than the rang of $G/[G,G]$. If you want to exclude the case where $H$ is abelian itself, just add two of the original generators to the commutator subgroup so the abelianization is of rank $2+\binom{k-1}{2}$.

For the modified version of (2) (see comments), it is still not true that $\mathrm{rank}(Z_H)\leq \mathrm{rank}(Z_G)$. Take $G$ as above; then $Z_G=[G,G]$ is free abelian of rank $\binom{k}{2}$. Take $H=\langle [G,G], x_1\rangle$, where $x_1$ is one of the free generators of $G$. Then $H$ is free abelian of rank $\binom{k}{2}+1$.

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 If G is nilpotent of class 2, then the third term of the lower central series is trivial, is it not? – Stuart May 19 2011 at 17:28 @Stuart: Oh, I see what the problem was; a typo. I meant "the third term of the lower central series of $F_k$; fixed. – Arturo Magidin May 19 2011 at 17:38 I didnt realize $Z_H \geq (H \cap Z_G)$. Then that means that $HZ_G/Z_G$ is actually a subgroup and everything is fine. Thanks!! – Stuart May 19 2011 at 17:42 So if I follow correctly, because of that, it follows that ($H \cap Z_G$) is normal in $H$ because it is a subgroup of the center? – Stuart May 19 2011 at 17:47 Umm... yeah, clearly because if it is in the center it commutes with everything... thanks again! – Stuart May 19 2011 at 17:51
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