# Abstract Algebra HW

1. Nov 14, 2012

### nateHI

1. The problem statement, all variables and given/known data
Suppose $N \lhd G$ and $K \vartriangleleft G$ and $N \cap K = \{e\}$. Show that if
$n \in N$and $k \in K$, then $nk = kn$. Hint: $nk = kn$ if and
only if $nkn^{-1}k^{-1} = e$.

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

3. 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.

2. Nov 15, 2012

### 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 $N \cap K$.

3. Nov 15, 2012

### 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 $k\in K$

4. Nov 15, 2012

### jbunniii

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?

5. Nov 15, 2012

### nateHI

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.

6. Nov 15, 2012

### jbunniii

OK, this is better. But the problem is that the n and k in this equation:
need not be the same as the n and k in this equation:
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.

7. Nov 15, 2012

### 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 :tongue2: which implies nkn^-1k^-1=e and I'm done. I just need to formalize that idea.

8. Nov 15, 2012

### jbunniii

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: $(nkn^{-1})k^{-1}$?

9. Nov 15, 2012

### HallsofIvy

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

10. Nov 15, 2012

### 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.

11. Nov 15, 2012

### 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.

12. Nov 15, 2012

### jbunniii

Looks good to me.