Non-Abelian Subgroup Produced by

[essie52]

This is not a homework question (although the answer will help answer a homework question). I know that a non-Abelian group can have both Abelian and non-Abelian subgroups but can a non-Abelian subgroup be produced by an Abelian group (or must the group be non-Abelian). Any thoughts appreciated. Thank you! E

 

[Dick]

This is not a homework question (although the answer will help answer a homework question). I know that a non-Abelian group can have both Abelian and non-Abelian subgroups but can a non-Abelian subgroup be produced by an Abelian group (or must the group be non-Abelian). Any thoughts appreciated. Thank you! E

State the definition of abelian. Then tell me how you could think an abelian group could have a nonabelian subgroup.

 

[essie52]

By Abelian I mean the elements of the group are commutative. For example, Z under multiplication is abelian since (a)(b) = (b)(a). Working backwards from my question, am I correct to say that the only way a non-Abelian group can produce an Abelian subgroup is if the subgroup is made of its inverses? With that reasoning, then a non-abelian subgroup would only be produced by a non-Abelian group.

 

[Office_Shredder]

Z under multiplication isn't a group.

If H is a subgroup of G, and G is abelian, and g and h are elements of H, why is gh=hg always?

Working backwards from my question, am I correct to say that the only way a non-Abelian group can produce an Abelian subgroup is if the subgroup is made of its inverse?

The subgroup is made up of what inverse?

 

[essie52]

"Z under multiplication isn't a group." Crap. Oops. That's what I get for multi-tasking. Correct, it is not a group (fails inverse test).