subgroups

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Sep 23rd 2009, 01:56 PM
Godisgood

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
Sep 23rd 2009, 07:13 PM
Chris L T521

Quote:

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 $e=e^n$ , so $A_n\neq\emptyset$ .

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

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

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

Does this make sense?
Sep 23rd 2009, 10:29 PM
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
Sep 23rd 2009, 11:19 PM
Swlabr

Quote:

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 $a$ and $a^{-1} \in H$ and $aa^{-1} \in H$ then as $aa^{-1}=1$ we have that $1 \in H$ ).

Firstly, note that your identity is $e^0$ . Secondly, does the question mention "finite" anywhere?
Sep 24th 2009, 08:15 AM
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
Sep 24th 2009, 08:34 AM
Swlabr

Quote:

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

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

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

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

For closure you need the abelian condition of $G$ .
Sep 24th 2009, 08:51 AM
Godisgood

I finally get it!!!
Thanks swlabr