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Abstract Algebra HW

by nateHI
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nateHI
#1
Nov14-12, 11:57 PM
P: 62
1. The problem statement, all variables and given/known data
Suppose [itex]N \lhd G [/itex] and [itex]K \vartriangleleft G[/itex] and [itex]N \cap K = \{e\}[/itex]. Show that if
[itex]n \in N [/itex]and [itex]k \in K[/itex], then [itex]nk = kn[/itex]. Hint: [itex]nk = kn[/itex] if and
only if [itex]nkn^{-1}k^{-1} = e[/itex].

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


3. The attempt at a solution
Let [itex]n_1,n_2\in N[/itex] and let [itex]k_1,k_2\in K[/itex]
Then
[itex](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[/itex]
But
[itex](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[/itex] where we used the fact that in this case, [itex]k_2 \notin N[/itex].
Therefore [itex]NK=KN[/itex] for all [itex]n\in N[/itex] and [itex]k\in K[/itex]

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

Please help.
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jbunniii
#2
Nov15-12, 12:07 AM
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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: [itex]n \in N[/itex] and [itex]k \in K[/itex]. Now consider the element [itex]nkn^{-1}k^{-1}[/itex]. The goal is to show that this equals [itex]e[/itex]. One promising way to do this would be to show that it is an element of [itex]N \cap K[/itex].
nateHI
#3
Nov15-12, 12:35 AM
P: 62
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 [itex]n_1,n_2\in N[/itex] and let [itex]k_1,k_2\in K[/itex]
Then
[itex]{\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[/itex]

But

[itex] {\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[/itex]

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

jbunniii
#4
Nov15-12, 12:49 AM
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Abstract Algebra HW

Quote Quote by nateHI View Post
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 [itex]n_1,n_2\in N[/itex] and let [itex]k_1,k_2\in K[/itex]
Then
[itex]{\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[/itex]
What does this string of equalities even mean? If we look at just the leftmost and rightmost expressions, you are asserting that
[tex]n_1 k_1 n_2 k_2 = NK[/tex]
The left hand side is an element, but the right hand side is a group. How can an element equal a group?
nateHI
#5
Nov15-12, 01:01 AM
P: 62
Quote Quote by jbunniii View Post
What does this string of equalities even mean? If we look at just the leftmost and rightmost expressions, you are asserting that
[tex]n_1 k_1 n_2 k_2 = NK[/tex]
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 [itex]n_1,n_2\in N[/itex] and let [itex]k_1,k_2\in K[/itex]
Then
[itex]{\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[/itex]

But

[itex] {\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[/itex]

I guess the flaw now is that we can't assume that [itex]nk_2=k_2n[/itex]. Well I'll just use your method cause this probably won't work.
jbunniii
#6
Nov15-12, 01:09 AM
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OK, this is better. But the problem is that the n and k in this equation:
[itex]{\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[/itex]
need not be the same as the n and k in this equation:
[itex] {\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[/itex]
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
#7
Nov15-12, 01:25 AM
P: 62
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
#8
Nov15-12, 01:59 AM
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Quote Quote by nateHI View Post
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 [itex]n \in N[/itex] and [itex]k \in K[/itex]. Now what can you say about the element [itex]nkn^{-1}k^{-1}[/itex]? Hint: what if you add some parentheses: [itex](nkn^{-1})k^{-1}[/itex]?
HallsofIvy
#9
Nov15-12, 06:52 AM
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Where have you used the fact that N and K are normal subgroups of G?
nateHI
#10
Nov15-12, 12:12 PM
P: 62
OK, let me try to make it a little more precise.

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

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

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

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



Am I done? It seems like there should be a more concise way to say it regardless.
nateHI
#11
Nov15-12, 12:31 PM
P: 62
Wait wait I get it now....

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

and

[itex](nkn^{-1})k^{-1}=e \implies nk=kn[/itex]


Took me a while but I got it.
jbunniii
#12
Nov15-12, 01:58 PM
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Quote Quote by nateHI View Post
Wait wait I get it now....

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

and

[itex](nkn^{-1})k^{-1}=e \implies nk=kn[/itex]


Took me a while but I got it.
Looks good to me.


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