The Abelianization of a Group

I know the definition of the Abelianization and the most direct implications of this new group, but what I don't understand are some of the broader implications. For instance, what can we say about the number of generators of the abelianization of a group with respect to the number of generators of that group? Also, one quick question: is it true that the abelianization of the direct product of groups is the direct product of the abelianization of the components (the direct product is finite)?

quasar987
Homework Helper
Gold Member
If a group G has S as a generating set, then for any normal subgroup N of G, SN={sN:s in S} is a generating set for G/N, and of course, |SN|<=|S|. If S is finite, then equality holds iff for all s, s' in S, $s^{-1}s'\notin N$.

This holds for any group. Were you looking for a sharper result specifically in the case where N=[G,G]=G' (the derived subgroup)?

As for your second question, it is not difficult and you should try answering it yourself by splitting the problem into simpler ones:
A) Show (GxH)/(NxM) ~ (G/N) x (H/M) for any group G,H and normal subgroups N,M (where ~ means isomorphic)
B) Show (GxH)'=G'xH' for any groups G,H
C) Conclude from this.

If a group G has S as a generating set, then for any normal subgroup N of G, SN={sN:s in S} is a generating set for G/N, and of course, |SN|<=|S|. If S is finite, then equality holds iff for all s, s' in S, $s^{-1}s'\notin N$.

This holds for any group. Were you looking for a sharper result specifically in the case where N=[G,G]=G' (the derived subgroup)?

As for your second question, it is not difficult and you should try answering it yourself by splitting the problem into simpler ones:
A) Show (GxH)/(NxM) ~ (G/N) x (H/M) for any group G,H and normal subgroups N,M (where ~ means isomorphic)
B) Show (GxH)'=G'xH' for any groups G,H
C) Conclude from this.

Thanks for the help; you've given me some good stuff to work out, and it looks like the result I was looking for is not only true, but shouldn't be too hard to prove.