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Abstract Algebra HW 
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#1
Nov1412, 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. 


#2
Nov1512, 12:07 AM

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PF Gold
P: 3,164

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].



#3
Nov1512, 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] 


#4
Nov1512, 12:49 AM

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P: 3,164

Abstract Algebra HW
[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? 


#5
Nov1512, 01:01 AM

P: 62

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. 


#6
Nov1512, 01:09 AM

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PF Gold
P: 3,164

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



#7
Nov1512, 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.



#8
Nov1512, 01:59 AM

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PF Gold
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#9
Nov1512, 06:52 AM

Math
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Thanks
PF Gold
P: 39,301

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



#10
Nov1512, 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. 


#11
Nov1512, 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. 


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