How Do Cosets Determine Group Element Relationships?

[fishturtle1]

Homework Statement

Let G be a group, and H a subgroup of G. Let a and b denote elements of G. Prove the following:

1. $Ha = Hb$ iff $ab^{-1} \epsilon H$.

Homework Equations

Let $e_H$ be the identity element of H.

The Attempt at a Solution

Proof: <= Suppose $ab^{-1} \epsilon H$. Then $abb^{-1} = a \epsilon Hb$. Since $H$ is a subgroup, $e_H \epsilon H$. So $e_Ha = a \epsilon Ha$. Since $Ha$ and $Hb$ share a common element, we must have $Ha = Hb$.

=> Suppose $Ha = Hb$. Since $e_ha = a \epsilon Ha$, we have $a \epsilon Hb$. So there must be some solution to $a = xb$ where $x \epsilon H$. Observe, $a = (ab^{-1})b$ so $ab^{-1} \epsilon H$ necessarily?

I'm not sure why $ab^{-1}$ would be the only solution. I think it has something to do with, if we wrote out the Cayley table for H, and there were two solutions for a = xb, then it wouldn't look right. Because in a Cayley table every element is in every row and every column. So we'd have an element missing in a row.

Also I didn't use the fact H is abelian so did I missing something there?

 

[fresh_42]

This is hard to correct, because it is so easy, that obvious and for reason is hard to distinguish.

Homework Statement

Let G be a group, and H a subgroup of G. Let a and b denote elements of G. Prove the following:

1. $Ha = Hb$ iff $ab^{-1} \epsilon H$.

Homework Equations

Let $e_H$ be the identity element of H.

The Attempt at a Solution

Proof: <= Suppose $ab^{-1} \epsilon H$. Then $abb^{-1} = a \epsilon Hb$.

Why? From $ab^{-1}\in H$ we get $ab^{-1}=h$ for some $h\in H$. Then $a=ab^{-1}b=hb\in Hb$. But why $abb^{-1}\,.$ And Abelian isn't an argument, for a) you haven't required it and b) it isn't required at all. It looks as if you used what you wanted to show.

Since $H$ is a subgroup, $e_H \epsilon H$. So $e_Ha = a \epsilon Ha$. Since $Ha$ and $Hb$ share a common element, we must have $Ha = Hb$.

=> Suppose $Ha = Hb$. Since $e_ha = a \epsilon Ha$, we have $a \epsilon Hb$.

I guess $e_h=e_H\,.$

So there must be some solution to $a = xb$ where $x \epsilon H$.

Yes, and therefore $ab^{-1}=(xb)b^{-1}=x\in H.$ I don't understand the rest.

Observe, $a = (ab^{-1})b$ so $ab^{-1} \epsilon H$ necessarily?

I'm not sure why $ab^{-1}$ would be the only solution. I think it has something to do with, if we wrote out the Cayley table for H, and there were two solutions for a = xb, then it wouldn't look right. Because in a Cayley table every element is in every row and every column. So we'd have an element missing in a row.

Also I didn't use the fact H is abelian so did I missing something there?

 

[fishturtle1]

Thank you for the reply, and sorry about all the typos.

Why? From ab−1∈Hab−1∈Hab^{-1}\in H we get ab−1=hab−1=hab^{-1}=h for some h∈Hh∈Hh\in H. Then a=ab−1b=hb∈Hba=ab−1b=hb∈Hba=ab^{-1}b=hb\in Hb. But why abb−1.abb−1.abb^{-1}\,. And Abelian isn't an argument, for a) you haven't required it and b) it isn't required at all. It looks as if you used what you wanted to show.

I meant to say, like you wrote: <= Suppose $ab^{-1} \epsilon H$. Then $(ab^{-1})b = a(b^{-1}b) = a \epsilon H$. Since $Ha$ and $Hb$ share an element other than $e_H$, it follows $Ha = Hb$.

For the 2nd part, => Suppose $Ha = Hb$. Then $a \epsilon Hb$. So there exists some $h \epsilon H$ such that $a = hb$. Multiplying by $b^{-1}$, we get $ab^{-1} = hbb^{-1} = h$. So $ab^{-1} \epsilon H$.

I don't understand the rest.

I think I was confused about if $x,y,z \epsilon G$ for some group G and $xy = xz$ then does $y = z$? And I see that it is true by multiplying $x^{-1}$ on both sides..

And Abelian isn't an argument, for a) you haven't required it and b) it isn't required at all.

I think because $abb^{-1} = a(bb^{-1}) = ae_H = a(b^{-1}b) = ab^{-1}b$

 

[Stephen Tashi]

I'm not sure why $ab^{-1}$ would be the only solution.

It's good to be suspicious about such things. Consult your course materials. In most presentations there is a theorem that says $a = xb$ has a unique solution for x.

In group theory, one can "do the same operation to both sides" of an equality and produce another equality, but the operation is limited to multiplying both sides by an element of the group. Explaining why this technique is reliable is an interesting exercise in logic! - perhaps an advanced exercise because it's hard to prove things that seem obvious.

Accepting the technique is valid, you could solve for $x$ by multiplying both sides of $a = xb$ on the right by $b^-1$.