Calculate the abeliazation of the group G=<a,b>

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SUMMARY

The abelianization of the group G=, where a and b are totally free and independent, can be calculated by first determining the commutator group [G,G]. The abelianization is then found by forming the quotient group G/[G,G]. The key insights include that the abelianization has no elements of finite order and can be represented as a free abelian group, which is a direct sum of copies of the integers. Furthermore, since a and b commute in the quotient group, they generate a free abelian group on one generator each.

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  • Understanding of group theory concepts, specifically commutator groups
  • Familiarity with quotient groups and their properties
  • Knowledge of free groups and free abelian groups
  • Basic grasp of generators in group theory
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Does anyone know how can I calculate the abeliazation of the group G=<a,b>
(when a and b are totally free and independent)
i.e. first of all, how do I calculate the commutator group [G,G]?
and then how do I find the quotient group: G/[G,G]

Thanks
 
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ibc said:
Does anyone know how can I calculate the abeliazation of the group G=<a,b>
(when a and b are totally free and independent)
i.e. first of all, how do I calculate the commutator group [G,G]?
and then how do I find the quotient group: G/[G,G]

Thanks

I would just count the number of generators in the abelianization then appeal to two facts: the abelianization has no elements of finite order and a free abelian group is a sum of copies of the integers.

Or you could directly show that a and b commute in the quotient group and that there are no other relations - e.g. each generate a free abelian group on one generator. For instance the powers of a and the powers of b are not commutators.
 

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