Abstract Algebra HW: Show nk=kn for N,K ∈ G

[nateHI]

Homework Statement

Suppose N \lhd G and K \vartriangleleft G and N \cap K = \{e\}. Show that if
n \in Nand k \in K, then nk = kn. Hint: nk = kn if and
only if nkn^{-1}k^{-1} = e.

Homework Equations

These "relevant equations" were not provided with the problem I'm just putting them here to make my solution more clear.
e=k_1^{-1}k_1
e=n_1^{-1}n_1

The Attempt at a Solution

Let n_1,n_2\in N and let k_1,k_2\in K
Then
(n_1)(k_1)(n_2)(k_2)=(n_1)(k_1)(n_2)e(k_2)=n_1(k_1n_2k_1^{-1})(k1k2)=n1NK=NK
But
(n_1)(k_1)(n_2)(k_2)=(n_1)(k_1)e(n_2)(k_2)=(n_1k_1n_1^{-1})(n_1n_2)k_2=KNk_2=Kk_2N=KN where we used the fact that in this case, k_2 \notin N.
Therefore NK=KN for all n\in N and k\in K

This seems correct to me but I didn't use the hint and my usage of N \cap K = \{e\} seems a little hand wavey.

Please help.

 

[jbunniii]

I didn't follow your proof. I think you are confusing things by using two elements from each subgroup. You only need the elements given in the problem statement: n \in N and k \in K. Now consider the element nkn^{-1}k^{-1}. The goal is to show that this equals e. One promising way to do this would be to show that it is an element of N \cap K.

 

[nateHI]

OK, I'll try your method next since I would like to know how to use the hint since the teacher is obviously trying to convey something he deems important. However, I modified my original method slightly and I think it works now. If someone has the time please tell if it does or how it doesn't work.

Let n_1,n_2\in N and let k_1,k_2\in K
Then
{\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)(n_2)e(k_2)=n_1(k_1n_2k_1^{-1})(k1k2)=n1NK=NK

But

{\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)e(n_2)(k_2)=(n_1k_1n_1^{-1})(n_1n_2)k_2=KNk_2=Kk_2N=KN

Where we used the facts that Nk_2=k_2N, n_1N=N and Kk_2=K. Therefore NK=KN for all n\in N and k\in K

 

[jbunniii]

OK, I'll try your method next since I would like to know how to use the hint since the teacher is obviously trying to convey something he deems important. However, I modified my original method slightly and I think it works now. If someone has the time please tell if it does or how it doesn't work.

Let n_1,n_2\in N and let k_1,k_2\in K
Then
{\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)(n_2)e(k_2)=n_1(k_1n_2k_1^{-1})(k1k2)=n1NK=NK

What does this string of equalities even mean? If we look at just the leftmost and rightmost expressions, you are asserting that
n_1 k_1 n_2 k_2 = NK
The left hand side is an element, but the right hand side is a group. How can an element equal a group?

 

[nateHI]

What does this string of equalities even mean? If we look at just the leftmost and rightmost expressions, you are asserting that
n_1 k_1 n_2 k_2 = NK
The left hand side is an element, but the right hand side is a group. How can an element equal a group?

Hmm, good point. I'll give this one last try using my method. Even if I don't get it this is good I feel like I'm learning a lot.

Here is my modification that I hope fixes it.

Let n_1,n_2\in N and let k_1,k_2\in K
Then
{\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)(n_2)e(k_2)=n_1(k_1n_2k_1^{-1})(k_1k_2)=n_1nk=nk

But

{\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)e(n_2)(k_2)=(n_1k_1n_1^{-1})(n_1n_2)k_2=knk_2=kk_2n=kn

I guess the flaw now is that we can't assume that nk_2=k_2n. Well I'll just use your method cause this probably won't work.

 

[jbunniii]

OK, this is better. But the problem is that the n and k in this equation:

{\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)(n_2)e(k_2)=n_1(k_1n_2k_1^{-1})(k_1k_2)=n_1nk=nk

need not be the same as the n and k in this equation:

{\bf (n_1)(k_1)(n_2)(k_2)}=(n_1)(k_1)e(n_2)(k_2)=(n_1k_1n_1^{-1})(n_1n_2)k_2=knk_2=kk_2n=kn

You have the right idea, but you are making it more complicated than it needs to be. Try similar manipulations with the idea I suggested earlier.

 

[nateHI]

I guess since the intersection of n and k is {e} for both of my n's of N and k's of K then intuitively nk=e and nk=e which implies nkn^-1k^-1=e and I'm done. I just need to formalize that idea.

 

[jbunniii]

I guess since the intersection of n and k is {e} for both of my n's of N and k's of K then intuitively nk=e and nk=e which implies nkn^-1k^-1=e and I'm done. I just need to formalize that idea.

I didn't follow your argument. You have n \in N and k \in K. Now what can you say about the element nkn^{-1}k^{-1}? Hint: what if you add some parentheses: (nkn^{-1})k^{-1}?

 

[HallsofIvy]

Where have you used the fact that N and K are normal subgroups of G?

 

[nateHI]

OK, let me try to make it a little more precise.

Since N is normal in G, any element k\in K which also happens to be in G since K is a subgroup (the fact that K is normal isn't important until the next step) of G we get knk^{-1}=n \implies nk=kn for all n \in N.

Since K is normal in G, any element n\in N which also happens to be in G since N is a subgroup (the fact that N is normal is only important in the last step) of G we get nkn^{-1}=k \implies kn=nk for all k \in K.

However, the intersection of N and K can only be {e} therefore nk=e=kn \implies nkn^{-1}k^{-1}=en^{-1}k^{-1}=ee=e.

Furthermore (nkn^{-1})k^{-1} \in K and n(kn^{-1}k^{-1}) \in N



Am I done? It seems like there should be a more concise way to say it regardless.

 

[nateHI]

Wait wait I get it now...

If (nkn^{-1})k^{-1} \in K and n(kn^{-1}k^{-1}) \in N where we have used the fact that N ad K are normal in G then (nkn^{-1})k^{-1}=e

and

(nkn^{-1})k^{-1}=e \implies nk=kn


Took me a while but I got it.

 

[jbunniii]

Wait wait I get it now...

If (nkn^{-1})k^{-1} \in K and n(kn^{-1}k^{-1}) \in N where we have used the fact that N ad K are normal in G then (nkn^{-1})k^{-1}=e

and

(nkn^{-1})k^{-1}=e \implies nk=kn


Took me a while but I got it.

Looks good to me.