The Abelianization of a Group: Implications and Proofs

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In summary, the abelianization of a group is the direct product of the abelianizations of the individual groups in the group.
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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.
 
  • #3
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.
 

1. What is the Abelianization of a group?

The Abelianization of a group, also known as the commutator subgroup, is a fundamental concept in group theory that involves finding the largest normal subgroup of a given group where all elements commute with each other.

2. Why is the Abelianization of a group important?

The Abelianization of a group is important because it helps to simplify and understand the structure of a group by reducing the non-commutative elements. It also has applications in other areas of mathematics, such as topology and algebraic geometry.

3. How is the Abelianization of a group calculated?

The Abelianization of a group is calculated by taking the quotient of the group by its commutator subgroup. This can be done by finding the commutator of all possible pairs of elements in the group and then taking the subgroup generated by these commutators.

4. Can the Abelianization of a group be trivial?

Yes, the Abelianization of a group can be trivial, meaning that the commutator subgroup is the identity element. This occurs when the group is already an abelian group, or when it has no non-commutative elements.

5. How does the Abelianization of a group relate to other group properties?

The Abelianization of a group is closely related to other group properties, such as the center, normal subgroups, and quotient groups. It can also be used to determine whether a group is simple or not, and to classify groups into different isomorphism classes.

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