# Thread: Isomorphic to the product of the subgroups H and K.

1. ## Isomorphic to the product of the subgroups H and K.

Here's the problem:

I cannot figure this out. So if H is not normal in G, does that mean that HxK is not isomorphic to G? Or could I define some other map that would be isomorphic?

2. Is H commutative?
Is K commutative?
Is HxK commutative?
Is G commutative?

3. Yes.
Yes.
Yes.
No.

I think I'm on the brink of seeing the solution. Are you saying that if HxK is commutative then G must also be commutative in order for HxK to be isomorphic to G? And if so, why is this?

4. Originally Posted by CropDuster
Yes.
Yes.
Yes.
No.

I think I'm on the brink of seeing the solution. Are you saying that if HxK is commutative then G must also be commutative in order for HxK to be isomorphic to G? And if so, why is this?
Two groups are isomorphic if they are the same `up to labelling'. They are essentially the same, just every element has a different name. For example, you can think of the integers as $\displaystyle \{\ldots -2, -1, 0, 1, 2, 3 \ldots\}$ under addition, or you can think of them as $\displaystyle \{\ldots a^{-2}, a^{-1}, 1, a, a^2 \ldots\}$ under multiplication. They are still the same group!

Basically, if $\displaystyle H \times K$ and $\displaystyle G$ are isomorphic then let $\displaystyle \phi:H\times K \mapsto G$ be the isomorphism. As $\displaystyle aba^{-1}b^{-1}=1$ for all $\displaystyle a, b \in H\times K$ we must have that $\displaystyle (aba^{-1}b^{-1})\phi=1$ in $\displaystyle G$. Aka, $\displaystyle (a\phi)(b\phi)(a^{-1}\phi)(b^{-1}\phi)=1$. As $\displaystyle \phi$ is onto, you are done.

(Note that $\displaystyle aba^{-1}b^{-1}=1 \Leftrightarrow ab=ba$)