# Thread: Proving that two groups are isomorphic

1. ## Proving that two groups are isomorphic

Hi.
I know that if I can show an isomorphism between two groups then that does it, but correct me if I'm wrong here (because really - that's only based on intuition, since we just started today to learn about isomorphism), most of the cases - it won't be easy finding a bijection between two groups.
So it'll be nice to have some more prespective and maybe some general guidelines that will make it easier to prove it.
Is there a general way to show that two groups are isomorphic to one another?

2. ## Re: Proving that two groups are isomorphic

Originally Posted by Stormey
I know that if I can show an isomorphism between two groups then that does it, but correct me if I'm wrong here (because really - that's only based on intuition, since we just started today to learn about isomorphism), most of the cases - it won't be easy finding a bijection between two groups.
So it'll be nice to have some more prespective and maybe some general guidelines that will make it easier to prove it.
Is there a general way to show that two groups are isomorphic to one another?
It is not enough to just find a bijection two groups operations must be preserved.
The key here is for the two groups to have the same structure.

Compare the Klein 4-group with the group on $\displaystyle \{0,1,2,3\}$ with operation of addition mod 4. Can you see that no isomorphism is possible, but because that have the same order there are $\displaystyle 24$ bijections?

3. ## Re: Proving that two groups are isomorphic

Yeah, I forgot that part about the function being homomorphism too.

4. ## Re: Proving that two groups are isomorphic

Often, what one does is the following:

1) One finds a surjective homomorphism $\displaystyle \phi:G \to G'$

2) One shows that $\displaystyle \text{ker}(\phi) = \{e_G\}$.

It is instructive to see what fails in this approach using Plato's example:

Suppose $\displaystyle G = \mathbb{Z}_4$ (under addition modulo 4), $\displaystyle G' = V = \{e,a,b,ab\}$, where $\displaystyle a^2 = b^2 = (ab)^2 = e$.

Since $\displaystyle G$ is cyclic, and any homomorphism satisfies:

$\displaystyle \phi(k) = \phi(1 + 1 +\cdots + 1)\ (k \text{ times})$

$\displaystyle = (\phi(1))^k$ it is enough to see where $\displaystyle \phi$ takes 1, a generator of $\displaystyle \mathbb{Z}_4$.

There are really only two cases:

$\displaystyle |\phi(1)| = 1$, in which case the homomorphism $\displaystyle \phi$ is trivial (and not surjective, the image is $\displaystyle \{e\}$).

$\displaystyle |\phi(1)| = 2$, in which case the image is of order 2 (and thus has just two elements, and thus cannot be surjective).

The conclusion is that there isn't ANY surjective homomorphism between the two groups, and thus no isomorphism (which would be a surjective homomorphism with trivial kernel).

5. ## Re: Proving that two groups are isomorphic

Great, thanks.