1. ## subgroups

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

2. Originally Posted by Godisgood
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
We use the one step subgroup test.

First note that $\displaystyle e=e^n$, so $\displaystyle A_n\neq\emptyset$.

Now, take $\displaystyle a^n,b^n\in A_n$. It follows then

$\displaystyle a^n\left(b^n\right)^{-1}=a^n\left(b^{-1}\right)^n=\left(ab^{-1}\right)^n$.

Since $\displaystyle ab^{-1}$ is in the right form, it follows that $\displaystyle A_n\leq G$

Does this make sense?

3. 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

4. Originally Posted by Godisgood
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 $\displaystyle a$ and $\displaystyle a^{-1} \in H$ and $\displaystyle aa^{-1} \in H$ then as $\displaystyle aa^{-1}=1$ we have that $\displaystyle 1 \in H$).

Firstly, note that your identity is $\displaystyle e^0$. Secondly, does the question mention "finite" anywhere?

5. 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

6. Originally Posted by Godisgood
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 $\displaystyle a$ is any element of the group, while the $\displaystyle n$ remains fixed.. For instance, take $\displaystyle G=C_8=\{1,g,g^2, \ldots, g^7\}$, with $\displaystyle g^8=1$. Let us look at this group when $\displaystyle n=2$, $\displaystyle A_2$. This is the set of squares of all elements of the group:

$\displaystyle A_2 = \{1^2=1, g^2, g^4, g^6 \}$.

The identity element is NOT $\displaystyle a^0$ - the $\displaystyle n$ is fixed. It is the elements of the group that we change. The identity occurs when $\displaystyle a=1$ as $\displaystyle 1^n=1 \forall n \in \mathbb{N}$.

If $\displaystyle g \in G$ then $\displaystyle g^{-1} \in G$. Thus $\displaystyle g^n \in A_n$ and $\displaystyle (g^{-1})^n = g^{-n} \in A_n$.

For closure you need the abelian condition of $\displaystyle G$.

7. I finally get it!!!
Thanks swlabr