My first time posting here! Wish all of you a merry Christmas! I am alone in my apartment, reviewing the homework for the semester and found some problems…

So we have an interesting fact: a subgroup of a finitely generated group is not necessarily finitely generated (a nice exercise by considering the commutator subgroup of the free group of two generators).

So my questions are:

- Show that if a finitely generated
*abelian*group is generated by elements, then any subgroup of is generated by elements. (This is a mid-term problem, and I forgot how to do it…) - Is the same statement true for finite but non-abelian group? (Open to me.)

Thanks!

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Oh I got (1). Sorry about that!

One can show that there exists a surjective homomorphism , either directly or from the universal property of free abelian groups. Then, if is any subgroup of , then is a subgroup of . Then we use the fact that any subgroup of must be isomorphic to where . Then must be generated by elements.

This argument does not apply to (2) because a surjective homomorphism from a non-abelian group to a subgroup needs not exist, say to , by order consideration.

wait, order of , and is a normal subgroup of …

If there’s such surjective homomorphism , we will draw a contradiction as follows.

has exactly one order 1 element and two order 3 element. For each transposition of (i.e. is of the form ), as , the order of must divide 2, so it must be 1 (since 3 does not divide 2). So we conclude that . As is generated by all the transpositions, must be trivial, which contradicts its surjectivity.

The answer to (2) is no in general.

Take , the permutation group of six elements, then it can be generated by two elements: and , this is quite clear because each transposition of the form can be generated by these two elements.

On the other hand, consider the subgroup generated by . We clearly have and , from this it is easy to see that , whose number of generators is at least 3.