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
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
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?
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
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 .