Question

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

Answer 1

=>
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?

Answer 2

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)

Answer 3

I don't know. Lol. This is abstract algebra?

Answer 4

yes

Answer 5

Hmm..I don't remember seeing this at all. Is this for part I or II?

Answer 6

its a 300 level group theory class. We don't have part 1 or 2. Its about cosets and fundamental homomorphism theorem.

Answer 7

I'm not quite sure how you get the first implication.

Answer 8

which part?

Answer 9

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 \(x\cdot a^n\in H\)?

Answer 10

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

Answer 11

I don't know what Im doing lol

Answer 12

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.

Answer 13

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.

Answer 14

Don't delete it. I might use some of it.

Answer 15

the book says Ha = Hb iff a*b^(-1) is in H is key

Answer 16

ok

Answer 17

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.

Answer 18

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.

Answer 19

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

Answer 20

how do we know these things?

Answer 21

ok the second one I get

Answer 22

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 \(H\).

Answer 23

so that is saying for an x in G we know there exists an a in G s.t. ga is in H?

Answer 24

I think I understand

Answer 25

That's basically it.

Answer 26

OK I got to go write it down and look at it for a bit, ty

Answer 27

Send me a message or reply if anything needs explaining. I've got to go in a few.

Answer 28

Im confused on " So g∈a k H for some k∈Z "

Answer 29

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?

Answer 30

ahh because G/H is a partition of G, so g has to be in some g'H where g' is in G

Answer 31

Yup. And we might as well call it gH instead of g'H since they're equal.

Answer 32

right. ok im picking up what your putting down :)

Answer 33

Hey @KingGeorge I'm confused on we got
\[H=Hxa^{k}\]

Answer 34

On how we got*

Answer 35

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

Answer 36

does that work?

Answer 37

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 \(xa^k\in H\).

Answer 38

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?

Answer 39

Well, the proof for that is fairly obvious since you multiplied by the same element x.

And I think your way also works.

Answer 40

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?

Answer 41

That's the one step I'm not entirely sure about. Give me a couple minutes to see if I can prove/disprove that.

Answer 42

Since \(a^na^{-n}=e\in H\), you know that \(a^nH=a^{-n}H\), so it does work.

Answer 43

And of course, \(a^{-n}=(a^{-1})^n\)

Answer 44

ty

Answer 45

You're welcome.

Answer 46

@kinggeorge the them says a*b^-1 off Ha=Hb did you use something different?

Answer 47

Theroem
* iif *

Answer 48

I used that theorem for the \(\Leftarrow\) direction, but not the other direction.

Answer 49

a*a^-1 implies Ha =Ha not Ha =Ha^-1

Answer 50

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
\[Ha=Ha\]

Answer 51

@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

Answer 52

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).

Answer 53

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