zzr0ck3r
  • zzr0ck3r
group theory: G is a group, H is a normal subgroup of G prove: G/H is cyclic iff there exists a in G s.t. for all x in G there exists n in Z s.t. x*a^n is in H
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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zzr0ck3r
  • zzr0ck3r
=> G/H is cyclic so H(ab) = H(ba) for a,b in G so (ab)(ba)^(-1) in in H (from theorem that states Ha=Hb iff ab^-1 is in H) so ab*a^(-1)*b^(-1) = h for some h in H since H is normal for any x in G we have x*h*x^(-1) is in H so x*(ab*a^(-1)*b(-1))*x^(-1) is in H and aba^(-1)b^(-1)x^(-1) is in G so we have shown what we need to show?
zzr0ck3r
  • zzr0ck3r
do I need to even do all this or can I say e is in H so e = x*a*a^(-1)*x^(-1) = x*(a*a^(-1)*x^(-1)) = a*g is in H where g = a*a^(-1)*x^(-1)
abb0t
  • abb0t
I don't know. Lol. This is abstract algebra?

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zzr0ck3r
  • zzr0ck3r
yes
abb0t
  • abb0t
Hmm..I don't remember seeing this at all. Is this for part I or II?
zzr0ck3r
  • zzr0ck3r
its a 300 level group theory class. We don't have part 1 or 2. Its about cosets and fundamental homomorphism theorem.
KingGeorge
  • KingGeorge
I'm not quite sure how you get the first implication.
zzr0ck3r
  • zzr0ck3r
which part?
KingGeorge
  • KingGeorge
Perhaps I'm missing something, but where have you shown that there exists an \(a\in G\) such that for all \(x\in G\) there exists an \(n\in\mathbb{Z}\) such that \(x\cdot a^n\in H\)?
zzr0ck3r
  • zzr0ck3r
e = x*a*a^(-1)*x^(-1) = x*(a*a^(-1)*x^(-1)) = a*g is in H where g = a*a^(-1)*x^(-1) for this one there exists a in G ((a*a^(-1)*x^(-1)) <-- this is my element in g) st for any x in G there exists n in Z (n=1) so x*(a*a^(-1)*x^(-1) = e is in H
zzr0ck3r
  • zzr0ck3r
I don't know what Im doing lol
KingGeorge
  • KingGeorge
Ah. That doesn't work. You're choosing a specific \(x\), not a general one. Give me a couple minutes to see if I can get something better.
zzr0ck3r
  • zzr0ck3r
ahh your right... Im going to delete all that stuff as not to confuse anyone else...lol. The wife just sait dinner is ready, ill be back in a bit.. sorry.
KingGeorge
  • KingGeorge
Don't delete it. I might use some of it.
zzr0ck3r
  • zzr0ck3r
the book says Ha = Hb iff a*b^(-1) is in H is key
zzr0ck3r
  • zzr0ck3r
ok
KingGeorge
  • KingGeorge
Since \(G\) is a group, we know that \(xg\in H\) for some \(g\in G\). We also know that \(G/H=\langle aH\rangle\) for some \(a\in G\). So \(g \in a^k H\) for some \(k\in\mathbb{Z}\). Then \(a^k H=g H\) since \(H\lhd G\). But since \(xg\in H\), we know that \(xg H=H=xa^k H\). Thus, \(xa^k\in H\). I think this works for the first direction.
KingGeorge
  • KingGeorge
Now for the other direction. Let \(a\in G\) such that for all \(x\in G\) there exists \(n\in\mathbb{Z}\) such that \(xa^n\in H\). Now notice that \(x^{-1}H=a^n H\) (since \(xa^n\in H\)). For simplicity, since \(x\) is general, let \(x^{-1}\) be our new \(x\). So \(x H=a^n H\) for all \(x\in G\), and some \(n\in\mathbb{Z}\). Thus, \(G/H=\langle aH\rangle\), so \(G/H\) is cyclic.
zzr0ck3r
  • zzr0ck3r
hmm, Since G is a group, we know that xg∈H for some g∈G . We also know that G/H=⟨aH⟩ for some a∈G
zzr0ck3r
  • zzr0ck3r
how do we know these things?
zzr0ck3r
  • zzr0ck3r
ok the second one I get
KingGeorge
  • KingGeorge
Well, the first we know since \(xx^{-1}=e\in H\). But we also know that \(xg\neq xh\) for all \(g\neq h\). So just by going through all of the possible \(g\in G\), \(xg\) is every possible element; including the ones in \(H\).
zzr0ck3r
  • zzr0ck3r
so that is saying for an x in G we know there exists an a in G s.t. ga is in H?
zzr0ck3r
  • zzr0ck3r
I think I understand
KingGeorge
  • KingGeorge
That's basically it.
zzr0ck3r
  • zzr0ck3r
OK I got to go write it down and look at it for a bit, ty
KingGeorge
  • KingGeorge
Send me a message or reply if anything needs explaining. I've got to go in a few.
zzr0ck3r
  • zzr0ck3r
Im confused on " So g∈a k H for some k∈Z "
zzr0ck3r
  • zzr0ck3r
so since xg is in H xg = h for some h so g = x^(-1)h so g is in one of the cosets? is that what that is saying?
zzr0ck3r
  • zzr0ck3r
ahh because G/H is a partition of G, so g has to be in some g'H where g' is in G
KingGeorge
  • KingGeorge
Yup. And we might as well call it gH instead of g'H since they're equal.
zzr0ck3r
  • zzr0ck3r
right. ok im picking up what your putting down :)
zzr0ck3r
  • zzr0ck3r
Hey @KingGeorge I'm confused on we got \[H=Hxa^{k}\]
zzr0ck3r
  • zzr0ck3r
On how we got*
zzr0ck3r
  • zzr0ck3r
so we know ab^(-1) is in H iff Ha = Hb so take x in G then Hx = Ha^n for some a in G and n in N and \[Ha^{n} = H(a^{-1})^{k}\] so \[Hx = H(a^{k})^{-1}\] so xa^k is in H
zzr0ck3r
  • zzr0ck3r
does that work?
KingGeorge
  • KingGeorge
We know that \(xgH=xa^kH\) since \(a^kH=gH\). But from our assumption, \(xg\in H\). So \(xgH=H\). Thus, \(xa^kH=H\) by the transitivity of equality. Therefore \(xa^k\in H\).
zzr0ck3r
  • zzr0ck3r
ok I hear ya, I just have never seen the prrof that aH=bH implies xaH=xbH how to you think the way I did it looks?
KingGeorge
  • KingGeorge
Well, the proof for that is fairly obvious since you multiplied by the same element x. And I think your way also works.
zzr0ck3r
  • zzr0ck3r
yeah after I said that I noticed, its just aH=bH implies a=bH so xa=xbH so xa is in xbH so xaH = xbH as far as mine, does a^nH = (a^-1)^kH for some n and k in Z?
KingGeorge
  • KingGeorge
That's the one step I'm not entirely sure about. Give me a couple minutes to see if I can prove/disprove that.
KingGeorge
  • KingGeorge
Since \(a^na^{-n}=e\in H\), you know that \(a^nH=a^{-n}H\), so it does work.
KingGeorge
  • KingGeorge
And of course, \(a^{-n}=(a^{-1})^n\)
zzr0ck3r
  • zzr0ck3r
ty
KingGeorge
  • KingGeorge
You're welcome.
zzr0ck3r
  • zzr0ck3r
@kinggeorge the them says a*b^-1 off Ha=Hb did you use something different?
zzr0ck3r
  • zzr0ck3r
Theroem * iif *
KingGeorge
  • KingGeorge
I used that theorem for the \(\Leftarrow\) direction, but not the other direction.
zzr0ck3r
  • zzr0ck3r
a*a^-1 implies Ha =Ha not Ha =Ha^-1
zzr0ck3r
  • zzr0ck3r
OK sorry I was on my tablet and this site is hard to use on it... \[Ha=Hb \iff a*b^{-1} \in H\] you have shown \[a*a^{-1} \in H\] and thus \[Ha=Ha\]
zzr0ck3r
  • zzr0ck3r
@Zarkon can you help on this one? This is what I have so we know ab^(-1) is in H iff Ha = Hb so take x in G then Hx = Ha^n for some a in G and some n in Z so Ha^n =H(a^−1)^k for some k in Z (this is the statement I am worried about, in other words I cant prove it:)) so Hx=H(a^k )^(-1) so xa^k is in H
KingGeorge
  • KingGeorge
I have a few things to say about that. First, I think that's a valid way of doing it. Second, you might want to say that \(Hx=Ha^n\) for some \(a\in G\) and \(n\in\mathbb{Z}\) because \(G/H\) is cyclic. Third, that statement you can't prove, is true if you let \(k=n\). Then since \((a^{-1})^n=(a^n)^{-1}\), you know that \(a^n\cdot (a^{-1})^n=e\), which is in \(H\) since \(H\) is a subgroup (and thus non-empty).
zzr0ck3r
  • zzr0ck3r
Yeah I just said since a^n is in G it has and inverse, and is then itself an inverse of some other element....and renamed is a^-1

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