most recent 30 from http://mathoverflow.net 2013-05-26T00:44:16Z http://mathoverflow.net/feeds/user/15228 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center Stuart 2011-05-19T16:22:28Z 2011-05-19T17:38:20Z

<p>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.</p> <p>A few questions:</p> <p>1) Suppose $G/Z_G$ $\simeq$ $\mathbb{Z}^k$ and $H/Z_H$ $\simeq$ $\mathbb{Z}^t$</p> <p>Is it always the case that $t \leq k$?</p> <p>2) What if $Z_G$ and $Z_H$ are replaced by $[G,G]$ and $[H,H]$ respectively?</p>

http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center/65457#65457 Stuart 2011-05-19T17:51:10Z 2011-05-19T17:51:10Z

Umm... yeah, clearly because if it is in the center it commutes with everything... thanks again!

http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center/65457#65457 Stuart 2011-05-19T17:47:17Z 2011-05-19T17:47:17Z

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?

http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center/65457#65457 Stuart 2011-05-19T17:42:26Z 2011-05-19T17:42:26Z

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!!

http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center/65457#65457 Stuart 2011-05-19T17:28:08Z 2011-05-19T17:28:08Z

If G is nilpotent of class 2, then the third term of the lower central series is trivial, is it not?

http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center Stuart 2011-05-19T17:24:37Z 2011-05-19T17:24:37Z

Sorry... I mean it's not necessarily a direct sum of subgroups.

http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center Stuart 2011-05-19T17:23:32Z 2011-05-19T17:23:32Z

@Carnahan: That is true. And originally I was mapping H to $HZ_G/Z_G$ + $Z_G \cap H$. But the problem is that this is not necessarily a subgroup so things get a bit ugly. I wanted to look for a different way to do it by taking it into the integers a straightforward way that would allow me to still call on the group structure. But maybe this is not doable.

http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center Stuart 2011-05-19T17:20:00Z 2011-05-19T17:20:00Z

@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.

http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center Stuart 2011-05-19T17:11:52Z 2011-05-19T17:11:52Z

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.

http://mathoverflow.net/questions/65455/quotient-of-subgroups-by-center Stuart 2011-05-19T17:02:45Z 2011-05-19T17:02:45Z

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 &lt; Z_G