The Abelianization of a Group: Implications and Proofs

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SUMMARY

The discussion centers on the Abelianization of a group, specifically addressing the relationship between the number of generators of a group and its Abelianization. It confirms that if a group G has a generating set S, then for any normal subgroup N of G, the set SN={sN:s in S} serves as a generating set for the quotient group G/N, with the cardinality |SN| being less than or equal to |S|. Additionally, it is established that the Abelianization of the direct product of groups is indeed the direct product of the Abelianizations of the individual groups, provided the direct product is finite.

PREREQUISITES
  • Understanding of group theory concepts, specifically Abelian groups and normal subgroups.
  • Familiarity with the notation and properties of quotient groups.
  • Knowledge of the derived subgroup and its implications in group theory.
  • Experience with isomorphism in the context of group products.
NEXT STEPS
  • Study the properties of normal subgroups in relation to group generators.
  • Learn about the derived subgroup and its role in the Abelianization process.
  • Explore the implications of the isomorphism (GxH)/(NxM) ~ (G/N) x (H/M).
  • Investigate the proof techniques for establishing the equality (GxH)'=G'xH' for groups G and H.
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Mathematicians, particularly those specializing in abstract algebra, group theorists, and students seeking to deepen their understanding of group properties and their implications in algebraic structures.

Newtime
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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)?
 
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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, [itex]s^{-1}s'\notin N[/itex].

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.
 
quasar987 said:
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, [itex]s^{-1}s'\notin N[/itex].

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.
 

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