# Elementary property of cosets

## 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
Mentor
2021 Award
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?

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
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