1. ## Abstract algebra

Hello all,

I have the following problem:

Suppose a is in G.
Let aG = {ag : g is in G}.
Show that aG=G.

It does not say anything about G, but I am assuming that G is a group and thus we have closure, associativity, an identity element and inverses for every element in G.

How do I start?

Thanks.

2. ## Re: Abstract algebra

Originally Posted by surjective
Suppose a is in G.
Let aG = {ag : g is in G}.
Show that aG=G.

It does not say anything about G, but I am assuming that G is a group and thus we have closure, associativity, an identity element and inverses for every element in G.
Assuming that $G$ is a group, you must show $aG\subset G~\&~G\subset aG$
1) If $x\in aG$ can you show that $x\in G~?$

2) If $y\in G$ can you show that $y\in aG~?$

3. ## Re: Abstract algebra

Here is what I have done:

Assuming that $\displaystyle x \in aG$ means that x can be written on the form $\displaystyle ag$ where $\displaystyle g \in G$. Since G is a group we have closure and therefore $\displaystyle ag \in G$.

Assuming that $\displaystyle y \in G$ means that $\displaystyle ay \in G$ due to closure. This means that ay is in aG since ay are the elements of aG. I am stuck here!!

Assistance would be appreciated.

4. ## Re: Abstract algebra

Originally Posted by surjective
Here is what I have done:
Assuming that $x \in aG$ means that x can be written on the form $ag$ where $g \in G$. Since G is a group we have closure and therefore $ag \in G$.
Do not use the [ tex] [ /tex] tags. Instead use \$\$ signs.
Yes you have shown that if $\in aG$ then $x\in G$

Now if $y\in G$ is it true that $y\in aG~?$

5. ## Re: Abstract algebra

If $y \in G$ then $ay \in G$ due to closure. Then $ay \in aG$?

6. ## Re: Abstract algebra

Originally Posted by surjective
If $y \in G$ then $ay \in G$ due to closure.
True. But how does that show that $y\in aG~?$

7. ## Re: Abstract algebra

Here I will finish this for you.
Suppose that $y\in G$. We know that $a^{-1}\in G$ so that means $a^{-1}y\in G$ WHY?
Thus that means $(\exists g'\in G)[a^{-1}y=g']$ But that implies that $y=ag'$ HOW?
If that shows that $y\in aG$ how have we proven that $aG=G~?$

8. ## Re: Abstract algebra

Yes I see.

Assume that $y \in G$. Assume also that $a \in G$. Since G is a group, then every element in G has an inverse which is also in G. Hence $a^{-1} \in G$.
Due to closure then $a^{-1}y \in G$. We may call this element $g'$ and thus $a^{-1}y=g' \in G$.
Applying the left cancellation law we get $a^{-1}y=g' \Leftrightarrow aa^{-1}y=ag' \Leftrightarrow y=ag'$.
Hence assuming that $y \in G$ then it has the form $ag'$ which are the elements of aG. Therefore $y \in aG$.

We have thus shown that $aG \subset G$ and $G \subset aG$ and therefore $G=aG$.

Thanks a lot.