First Isomorphism Theorem

mykayla10

1. The problem statement, all variables and given/known data
Suppose H is a normal subgroup G and L is a subgroup of K. Then (G x K)/(H x L) is isomorphic to (G/H) x (K/L)


2. Relevant equations



3. The attempt at a solution
I know that I have to use the First Isomorphism Theorem, but in order to do that I need some function phi. I am having a really difficult time finding a function from (G x K) to (G/H)x(K/L). If I have this I am almost certain I can complete the proof with the First Isomorphism Theorem.

Stephen Tashi

Does the problem say "L is a subgroup of K"? It's not necessarily a normal subgroup, correct?

mykayla10

It says that L is a normal subgroup of K.

Stephen Tashi

That's good. You have already studied the picture where there is homomorphism from G to some other group (which could also be G) and map phi1 from G to G/H.

Livewise you can have a picture where there is a map from K to some other group and a map phi2 from K to K/L. The group G X K is just ordered pairs of elements from the two groups that look like {g, k}. You can use phi1 to map the g-element into G/H and the phi2 map to map the k-element into K/L. That defines a map from GxH to G/H x K/L.

mykayla10

you are allowed to use two different phis??

Stephen Tashi

You define a single map, often called phi1 x phi2, whose definition involves using both phi1 and phi2. So I suppose the answer is "yes, you are allowed to use two different phis", but you must word it so you are defining a single map by using them.

mykayla10

Ok, so that makes sense. Now I am lost as to what the two phis should look like. phi1: G ->G/H and phi2: K -> K/L. However, I don't know what these look like. I am so bad at proofs.

Stephen Tashi

Read what your text says about G/H as a group. It might say that the elements of G/H are cosets of H or it might say thay are "coset representatives" or something like that. Whatever the terminology, to get a homomorphism phi1 from G to G/H, map the element G to the coset (or coset representative of) gH.