# Abelian subgroups proof

• Apr 22nd 2014, 08:42 PM
Educated
Abelian subgroups proof
Quote:

If A is an abelian group with $A \unlhd G$ and B is any subgroup of G, prove that $A \cap B \unlhd AB$
So I'm guessing AB is just defined as $AB : = \{ ab \, \, | \, \, a \in A, b \in B \}$

I started off with:

Let $x \in A \cap B, a \in A, b \in B$
We must show that $(ab) x (ab)^{-1} \in A \cap B$

$(ab) x (ab)^{-1} = ab x b^{-1} a^{-1}$

And now here I'm stuck...
• Apr 23rd 2014, 11:24 AM
johng
Re: Abelian subgroups proof
Hi,
My earlier post on exponential notation for conjugation is used here. I'm a firm advocate of this notation; it makes lots of proofs much easier.

http://i60.tinypic.com/dvg5j.png
• Apr 24th 2014, 06:56 PM
Educated
Re: Abelian subgroups proof
Thanks, I finally got it after reading it 3 or 4 times :p
• Apr 30th 2014, 02:34 AM
Educated
Re: Abelian subgroups proof
Actually reading that again, I still don't get how it's a subgroup...

So I realised there's an easier way to prove that it normal, but what about proving that it is an actual subgroup of AB, ie. $A \cap B \leq AB$

Also, here's my reasoning for a normal subgroup:
Since A and B are subgroups, $A \cap B$ is also a subgroup. In particular since A is abelian and $A \cap B \leq A$ then $A \cap B$ is also abelian, hence it is also a normal subgroup of G, ie. $A \cap B \unlhd G$

This means $gxg^{-1} \in A \cap B \, \, \forall \, \, x \in A \cap B, \, \, \forall \, \, g \in G$
Since $AB \subseteq G$ then
$gxg^{-1} \in A \cap B \, \, \forall \, \, x \in A \cap B, \, \, \forall \, \, g \in AB \subseteq G$

Would that work?
• Apr 30th 2014, 10:01 AM
johng
Re: Abelian subgroups proof
• Apr 30th 2014, 01:34 PM
Deveno
Re: Abelian subgroups proof
Suppose we have:

$abxb^{-1}a^{-1}$ with $a \in A,b \in B, x\in A \cap B$.

Since $AB$ is a group, $AB = BA$, so we can write this as:

$b'a'xa'^{-1}b'^{-1}$.

Now $A$ is abelian, and thus $a'xa'^{-1} = x$, so $b'a'xa'^{-1}b'^{-1} = b'xb'^{-1} \in B$

Also, $A \unlhd G$, so since $b' \in B \leq G$, $b'xb'^{-1} \in A$, so $abxb^{-1}a^{-1} = b'a'xa'^{-1}b'^{-1} \in A\cap B$

Proof that $AB$ is a subgroup iff $AB = BA$:

Suppose $AB$ is a group. Then, by closure, $abab = a'b'$ for any $a \in A,b \in B$ and some $a' \in A, b' \in B$.

Thus $ba = (a^{-1}a')(b'b^{-1}) \in AB$, so $BA \subseteq AB$.

Also, since $AB$ is a group, for $ab \in AB$ we have $(ab)^{-1} \in AB$, so $ab = (a'b')^{-1}$ for some $a' \in A, b' \in B$, thus $ab = (a'b')^{-1} = b'^{-1}a'^{-1} \in BA$, so $AB \subseteq BA$.

Now suppose $AB = BA$. We need to show closure and existence of inverses (that $e \in AB$ is trivial).

So take any product $aba'b'$. Since $BA = AB$, we can write $ba' = a''b''$, so $aba'b' = (aa'')(b''b') \in AB$, which shows closure.

Now if $ab \in AB$, then $(ab)^{-1} = b^{-1}a^{-1} \in BA = AB$, which shows $AB$ contains all inverses, and is thus a group.
• May 2nd 2014, 09:53 AM
johng
Re: Abelian subgroups proof
Hi,
Here's some more discussion that I hope you find helpful:

http://i59.tinypic.com/34hifpj.png