If G is an abelian group and n>1 an integer, let A_n = { a^n, a is an element of G}. Prove that A_n is a subgroup of G

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- September 23rd 2009, 02:56 PMGodisgoodsubgroups
If G is an abelian group and n>1 an integer, let A_n = { a^n, a is an element of G}. Prove that A_n is a subgroup of G

- September 23rd 2009, 08:13 PMChris L T521
- September 23rd 2009, 11:29 PMGodisgood
No not really.

What I tried doing was showing that A_n is closed, associative, has an identity element and an inverse element.

My problem was I got an identity element e = a^0

and inverse = a^-n.. I suppose both of them are wrng because th question says n>1....what do you think I did wrong..

Thnx - September 24th 2009, 12:19 AMSwlabr
To prove something is a subgroup you merely need to show Closure and that Inverses exist. You get associativity for free because you are in a group (think about it...), and the identity because inverses exist (if and and then as we have that ).

Firstly, note that your identity is . Secondly, does the question mention "finite" anywhere? - September 24th 2009, 09:15 AMGodisgood
No the question does not mention infinite..

Alo bare with me but am still confused as to why the identity element is e^0

I though a * e = a

and in this case since A_n = a^n

so is a ^n * e = a^n

that is how I got e = a^0

what am doing wrong

Thanks for your help so far - September 24th 2009, 09:34 AMSwlabr
The element is any element of the group, while the remains fixed.. For instance, take , with . Let us look at this group when , . This is the set of squares of all elements of the group:

.

The identity element is NOT - the is fixed. It is the elements of the group that we change. The identity occurs when as .

If then . Thus and .

For closure you need the abelian condition of . - September 24th 2009, 09:51 AMGodisgood
I finally get it!!!

Thanks swlabr