# How to find a proper mapping in order to prove isomorphism of groups

• Sep 28th 2010, 03:05 AM
rubix
How to find a proper mapping in order to prove isomorphism of groups
as the title says, given two groups, and if i'm asked to prove they're isomorphic...how do i find the mapping that satisfies? is there a general way of doing this? if not, what kind of thought process you do?

ex: show isomorphism from nZ to mZ
n and m are some integers
nZ and mZ are collection of multiples of n and m respectively
• Sep 28th 2010, 03:21 AM
Swlabr
Quote:

Originally Posted by rubix
as the title says, given two groups, and if i'm asked to prove they're isomorphic...how do i find the mapping that satisfies? is there a general way of doing this? if not, what kind of thought process you do?

ex: show isomorphism from nZ to mZ
n and m are some integers
nZ and mZ are collection of multiples of n and m respectively

I'm afraid to say that there is no general rule'. I mean, for your example I would prove that every subgroup of $\displaystyle \mathbb{Z}$ is isomorphic to $\displaystyle \mathbb{Z}$ (can you see why this is sufficient?). But if I was to prove that $\displaystyle D_4$, the dihedral group of order 4, was isomorphic to the Klein 4-group I would point out that it is not cyclic, and there are only two groups of order 4, one of which is cyclic, the other if the Klein 4-group.

However, when trying to find an isomorphism there are a few things to note,

If you map to the generators then your homomorphism will be surjective,

If the kernel is trivial then your homomorphism will be injective,

Try to find a common' group which both groups are isomorphic to (as in the example),

Often there exists some well-known function which actually turns out to be an isomorphism of groups (for example, $\displaystyle (\mathbb{R}, +) \cong (\mathbb{R}^{+}, \times)$, and the isomorphism is given by $\displaystyle e^x$).

A homomorphism is an isomorphism if and only if there exists an inverse function. Sometimes this can be easily found.
• Sep 28th 2010, 07:15 PM
rubix
transitive property? if subset of Z is isomorphic to Z then nZ and mZ are isomorphic?

but how would one actually go about proving isomorphism? in particular i'm thinking what the mapping would be?
• Sep 29th 2010, 12:10 AM
Swlabr
Quote:

Originally Posted by rubix
transitive property? if subset of Z is isomorphic to Z then nZ and mZ are isomorphic?

but how would one actually go about proving isomorphism? in particular i'm thinking what the mapping would be?

Well, I'll leave you to figure the isomorphism out. But as a helping hand, and another tip, remember that a homomorphism is defined precisely by where the generators are mapped to.

So, for example, you know that if $\displaystyle H \leq \mathbb{Z}$ then $\displaystyle H=<i>$ for some $\displaystyle i \in \mathbb{Z}$, by the Euclidean algorithm. So, $\displaystyle \phi: \mathbb{Z} \rightarrow H$ defined by $\displaystyle 1 \mapsto i$ defines an isomorphism (can you see why it defines an isomorphism?).