# Isomorphism between HK and H x K

1. Mar 8, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
H and K are normal subgroups of G such that the intersection of H and K is the identity. Also, G = HK = {hk | h in H and k in K}. Find an isomorphism between G and H x K

2. Relevant equations

3. The attempt at a solution

I was thinking that an isomorphism could be $\mu : G \rightarrow H \times K$ where $\mu(g) = \mu(hk) = (h, k)$. I think that this will work, and I just want some verification.

2. Mar 9, 2017

### Dick

It's easy to show if $G$ is commutative, right? Can you show it without assuming that?

Last edited: Mar 9, 2017
3. Mar 9, 2017

### Mr Davis 97

Yes. Since both h and k are normal and since both h and k are in G, they commute with each other.

4. Mar 9, 2017

### Dick

If you are quoting a theorem that says that, then it's wrong. Normality only tells you e.g. $hkh^{-1}=k'$ where $k'$ is in $K$. You need to use another premise.

5. Mar 9, 2017

### Mr Davis 97

I am not sure. I thought that normality said that hg = gh, but since all the k are in hk = kh

6. Mar 9, 2017

### Dick

No, that's not what normality means. Might be time to look up the definition...

7. Mar 9, 2017

### Mr Davis 97

In my textbook: "A subgroup H of a group G is normal of its left and right cosets coincide, that is, if gH = Hg for all g in G."

So we know that gH =Hg for all g and gK = Kg for all g.

8. Mar 9, 2017

### Staff: Mentor

Yes, but this doesn't mean elementwise, therefore it's only $gh=h'g$ and $gk=k'g$.

9. Mar 9, 2017

### Mr Davis 97

I am not sure how to proceed, because it seems that the statement of normality involves cosets, but I am trying to find a statement involving just elements, hk = kh

10. Mar 9, 2017

### Mr Davis 97

Could I do soomething like $hk = h(h^{-1} k h) = kh$?

edit: nevermind

11. Mar 9, 2017

### Staff: Mentor

You don't need to and won't find one. Whether you write it as cosets $gH=Hg$ or elementwise $gh=hg'$ doesn't matter.
The point is: Why is $HK$ a group? Or: How do you manage to sort elements of the kind $hkh'k'$ such that you can separate them into $H \times K$, i.e. pairs $(h'',k'')\,$?

12. Mar 9, 2017

### Mr Davis 97

Well, since e is in both H and K e is in HK.
We need to show that (hk)(h'k') is also in HK. To do this we would need to have commutativity so that (hk)(h'k') = h(kh')k' = h(h'k)k' = (hh')(kk'). But I haven't shown commutativity yet

13. Mar 9, 2017

### Staff: Mentor

You don't need commutativity. Only associativity and $kh'=h''k$. It is important to end up in the groups $H$ and $K$, not that the elements are invariant. After that you have to show that $\mu$ is surjective (obvious) and injective (what do you need here to use?) and a group homomorphism (which is basically the same as to show that $HK$ is a group.)

14. Mar 9, 2017

### Mr Davis 97

Take any two elements of HK, hk and h’k’. Then, we want to show that (hk)(h’k’) is in HK. We note that K is normal, which implies that for kh’, there is some k’’ in K such that kh’=h’k’’ since for all a in G, aK=Ka. Thus,
(hk)(h’k)’=h(kh’)k’=h(h’k’’)k’=hh’k’’k’. Then since hh’ is in H and k’’k’ is in K, we have that hkh’k’ is in HK.

So will my definition of $\mu$ suffice if I show that it is injective and a homomorphism?

Also, don't I need to eventually show that h and k commute? I am still not sure how to show that

15. Mar 9, 2017

### Dick

Yes, I think you do need to show $hk=kh$. I'll give you a big hint. Use that the intersection of $H$ and $K$ is the identity.

16. Mar 9, 2017

### Mr Davis 97

Let $g = hkh^{-1} k^{-1}$. Since $hkh^{-1} \in K$, $g \in K$. Since $kh^{-1} k^{-1} \in H$, we have that $g \in H$. So g is in the intersection of H and K. But that is just the identity, so $hk h^{-1} k^{-1} = e$ and $hk = kh$.

Now do I use this to show that $\mu (g) = \mu(hk) = (h, k)$ is a homomorphism?

17. Mar 9, 2017

### Staff: Mentor

Yes, right.
It's quite obvious, and if you had written it down, it would have been basically of the same length as your question. Only injectivity needs an argument, not really a proof, but a reason.

18. Mar 9, 2017

### Dick

Well, what would showing $\mu$ is a homomorphism require you to prove??

19. Mar 9, 2017

### Mr Davis 97

I know that I would use commutativity to show to show that it's a homomorphism, and injectivity is easy to show.
I'm wondering though whether mu is well-defined. It seems like a strange definition because the image does not involve g but rather h and k. Is the function well-defined precisely because each g is uniquely represented as a product hk for some h and some k? Also, if I were to use the function, given a g in G how would I know how to decompose it into the product hk in order to use the function?

20. Mar 9, 2017

### Staff: Mentor

The general case is $H/(H\cap K) \cong HK/K$ for subgroups $H,K$ of $G$ from which $K$ is normal.
In the given case this translates to $H \cong H/\{1\} =H/(H \cap K) \cong HK/K \cong (H \times K)/K \cong G/K \cong H$ and is the reason, why $H \cap K = \{1\}$ is important to answer those questions.
What is $G$ here? The product of subgroups $HK$ or the product $H\times K$? Which one do you mean?
In the latter case, it's obvious, and in the former, that's what you have shown: that you can sort the elements. So again, what is $G$ in your question?