# Proving that two groups are isomorphic

• Dec 13th 2013, 09:59 AM
Stormey
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?

• Dec 13th 2013, 11:05 AM
Plato
Re: Proving that two groups are isomorphic
Quote:

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 $\{0,1,2,3\}$ with operation of addition mod 4. Can you see that no isomorphism is possible, but because that have the same order there are $24$ bijections?
• Dec 13th 2013, 11:52 AM
Stormey
Re: Proving that two groups are isomorphic
Yeah, I forgot that part about the function being homomorphism too.
• Dec 16th 2013, 12:06 PM
Deveno
Re: Proving that two groups are isomorphic
Often, what one does is the following:

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

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

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

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

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

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

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

There are really only two cases:

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

$|\phi(1)| = 2$, 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).
• Dec 20th 2013, 01:29 PM
Stormey
Re: Proving that two groups are isomorphic
Great, thanks.