Abstract Algebra - isomorphism question

[nebbish]

Abstract Algebra -- isomorphism question

If N, M are normal subgroups of G, prove that NM/M is isomorphic to N/N intersect M.

That's how the problem reads, although I am not sure how to make the proper upside-down cup intersection symbol appear on this forum. Or how to make the curly "=" sign for an isomorphism.

First of all, I am uncertain whether in fact N/N intersect M is meant to read N/(N intersect M), but I am going on that assumption. Otherwise, N/N would just be N, wouldn't it?

The collection of groups given by NM/M can be found by taking group M and multiplying by members of N and M: Mnm. The groups are normal, so this can be rewritten as
nMm. The expression Mm = M since a group multiplied by one of its own members is itself. If we wish, we can move the n so that we end up with a simple right coset: Mn.

Basically then it seems to me that NM/M = the collection of right cosets Mn, where n is an element of N. If n is in the intersection of M and N then Mn = M. However, I am uncertain how to proceed if n is not in the intersection of M and N. Perhaps that does not require simplification. In any case, I am pretty mystified on how to proceed with the other expression of interest, N/N intersect M.

 

[StatusX]

If N, M are normal subgroups of G, prove that NM/M is isomorphic to N/N intersect M.

That's how the problem reads, although I am not sure how to make the proper upside-down cup intersection symbol appear on this forum. Or how to make the curly "=" sign for an isomorphism.

To make mathematical expressions, use tex brackets. For example:

$$NM/M \cong N/(N \cap M)$$

You can click on that to see the code, and to get a link to a page with more info.

First of all, I am uncertain whether in fact N/N intersect M is meant to read N/(N intersect M), but I am going on that assumption. Otherwise, N/N would just be N, wouldn't it?

That's a good assumption, although N/N would be the trivial group with a single element.

The collection of groups given by NM/M can be found by taking group M and multiplying by members of N and M: Mnm. The groups are normal, so this can be rewritten as
nMm. The expression Mm = M since a group multiplied by one of its own members is itself. If we wish, we can move the n so that we end up with a simple right coset: Mn.

Basically then it seems to me that NM/M = the collection of right cosets Mn, where n is an element of N. If n is in the intersection of M and N then Mn = M. However, I am uncertain how to proceed if n is not in the intersection of M and N. Perhaps that does not require simplification. In any case, I am pretty mystified on how to proceed with the other expression of interest, N/N intersect M.

I'm having a hard time understanding what you mean here. MN is the set of elements of the form mn, where m is in M and n is in N. This can be looked at as the union of the left cosets of N containing elements in M, ie, the nM, or as the union of the Nm, although for a single n or m, nM and Nm are not, in general, subgroups.

Forget about simplifying the expressions, that's probably not going to be possible. Focus on constructing an isomorphism. A general element in the group on the LHS is of the form mnM, while an element from the group on the RHS is of the form nM$\cap$N. To what element in the second group might you send mnM? Can you show the resulting map is an isomorphism?

 

[nebbish]

Thank you for your swift reply. Actually, we are using right cosets as standard in this course. So I would expect NM/M to be composed of groups of the form Mnm. However, because M is normal I believe we can turn this expression into just Mn.

That is, Mnm = (Mn)m = (nM)m = n(Mm) = nM = Mn.

That's about as far as I got when I posted here. It seems that you don't find this result of significance.

To continue with that result I would have:

I guess I would then take Mn --> $$(N \cap M)$$n, I suppose.
If $$(N \cap M)$$ were normal I might do the following:
Let n and r be elements of N.
Let U = $$(N \cap M)$$.

Then f((Mn)(Mr)) = f(Mnr) = Unr = UUnr = UnUr = f(Mn)f(Mr).

If that is the correct track, I would need to show f is one-to-one and onto (a bijection).
I would also need to show $$(N \cap M)$$ is normal. Any hints on proving those things would be appreciated.

Also if I am completely spinning out of control I would appreciate correction on that. Thanks.

 

[nebbish]

Actually, I guess proving the normality of $$(N \cap M)$$ is trivial, isn't it? When I learn more tex I might put it here.

But basically if r is in the intersection then xr(x-inverse) is in N by the normality of N and in M by the normality of M. Then the intersection itself must be normal.

 

[StatusX]

So I would expect NM/M to be composed of groups of the form Mnm.

As I said before, that is not a group.

However, because M is normal I believe we can turn this expression into just Mn.

That is, Mnm = (Mn)m = (nM)m = n(Mm) = nM = Mn.

That's about as far as I got when I posted here. It seems that you don't find this result of significance.

Sorry, I missed that the first time through. That is correct (another way to see this is to note that in G/M, Mm is the identity, so Mnm=MnMm=Mn), and it is significant.

To continue with that result I would have:

I guess I would then take Mn --> $$(N \cap M)$$n, I suppose.
If $$(N \cap M)$$ were normal I might do the following:
Let n and r be elements of N.
Let U = $$(N \cap M)$$.

Then f((Mn)(Mr)) = f(Mnr) = Unr = UUnr = UnUr = f(Mn)f(Mr).

If that is the correct track, I would need to show f is one-to-one and onto (a bijection).
I would also need to show $$(N \cap M)$$ is normal. Any hints on proving those things would be appreciated.

Also if I am completely spinning out of control I would appreciate correction on that. Thanks.

There's one small problem. You need to make sure that the map is well defined. That is, if Mn₁m₁ = Mn₂m₂, does it follow that Un₁=Un₂?

Everything else seems fine. Note that a homomorphism is an injection iff f(g)=e only for g the identity. And it is surjective if you can find an element to map to any element in the target. You should be able to prove both these facts without too much trouble.