Isomorphism between HK and H x K

[Mr Davis 97]

Homework Statement

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

Homework Equations

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.

 

[Dick]

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

 

[Mr Davis 97]

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

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

 

[Dick]

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

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.

 

[Mr Davis 97]

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

 

[Dick]

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

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

 

[Mr Davis 97]

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

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.

 

[fresh_42]

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.

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

 

[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

 

[Mr Davis 97]

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

edit: nevermind

 

[fresh_42]

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

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 $(h'',k'')\,$?

 

[Mr Davis 97]

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 $(h'',k'')\,$?

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

 

[fresh_42]

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

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.)

 

[Mr Davis 97]

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.)

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

 

[Dick]

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

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.

 

[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?

 

[fresh_42]

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 $hk = kh$.

Yes, right.

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

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.

 

[Dick]

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?

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

 

[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?

 

[fresh_42]

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?

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.

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?

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?