# Thread: Left and Right cosets.

1. ## Left and Right cosets.

Let H be a subgroup of G and suppose that the number n of right cosets of H is G is finite. Prove that there are exactly n left cosets of H in G.

(Hint: Consider the set of inverses of the elements in a right coset).
Let $|G:H|=|G|/|H|=n$ for $n \in \mathbb{N}$.

Let $H=\{h^0, \ldots , h^{n-1} \}$ for some $h \in H$.

Then $Hg=\{h^0g, \ldots , h^{n-1}g \}$ for $g \in G$.

$(Hg)^{-1}=\{(h^0g)^{-1}, \ldots , (h^{n-1}g)^{-1} \}$

This is equal to $(Hg)^{-1}=\{g^{-1}(h^0)^{-1}, \ldots ,g^{-1}(h^{n-1})^{-1} \}$________________(*)

Since inverses are unique, and $H$ is a group (closure axiom) we have thatb $$(h^k)^{-1}=h^i$$ for $0 \leq k \leq n-1$ for some $0 \leq i \leq n-1$.

Hence we have $(Hg)^{-1}=\{g^{-1}h^0, \ldots , g^{-1}h^{n-1} \} =g^{-1}H$

Since $|G:H|=n$, we have exactly $n \ g^{-1}$'s to multiply $H$ by.

Thus, we have $n$ left cosets of $H$ in $G$.

I think this is right (can someone check =S), but step (*) applies only to multiplicative groups. Is it a fact that addition is always commutative?

2. Originally Posted by Showcase_22
Let $|G:H|=|G|/|H|=n$ for $n \in \mathbb{N}$.

Let $H=\{h^0, \ldots , h^{n-1} \}$ for some $h \in H$.

Uh? You can't assume this unless you know $H$ is cyclic (which isn't a given) AND $h$ is one of its generators...

Then $Hg=\{h^0g, \ldots , h^{n-1}g \}$ for $g \in G$.

$(Hg)^{-1}=\{(h^0g)^{-1}, \ldots , (h^{n-1}g)^{-1} \}$

This is equal to $(Hg)^{-1}=\{g^{-1}(h^0)^{-1}, \ldots ,g^{-1}(h^{n-1})^{-1} \}$________________(*)

Since inverses are unique, and $H$ is a group (closure axiom) we have thatb $(h^k)^{-1}=h^i$ for $0 \leq k \leq n-1$ for some $0 \leq i \leq n-1$.

Hence we have $(Hg)^{-1}=\{g^{-1}h^0, \ldots , g^{-1}h^{n-1} \} =g^{-1}H$

Since $|G:H|=n$, we have exactly $n \ g^{-1}$'s to multiply $H$ by.

Thus, we have $n$ left cosets of $H$ in $G$.

I think this is right (can someone check =S), but step (*) applies only to multiplicative groups. Is it a fact that addition is always commutative?
.

They gave you a big hint and that's what you should try: assume $X_L\,,\,\,X_R$ represent the sets of left and right cosets of $G\,\,\,in\,\,\,H$ , and define

$f: X_L\rightarrow X_R \,\,\,by\,\,\,f(gH):=Hg^{-1}$ . Well, now prove this function is a bijection.

Tonio

3. I got caught up on the question before that was about cyclic groups!

As for the function being a bijection:

We have $n$ lots of $gH$. We also have $n$ lots of $Hg^{-1}$ (proof as follows):

Since we have $n$ lots of $g$, we have at most $n$ lots of $Hg^{-1}$. We will only have less if, for some $k \in G$, $Hg^{-1}=Hk^{-1} \Rightarrow \ g^{-1}=k^{-1} \Rightarrow \ g=k$ (since inverses are unique).
Therefore we have $n$ lots of $Hg^{-1}$.

Also, since inverses are unique, $g^{-1}$ is unique to a $g$. Therefore the function is injective.

We have a mapping from a set of $n$ elements to a set of $n$ elements which is injective. By the pigeonhole principle, the function is a bijection.

4. Originally Posted by Showcase_22
I got caught up on the question before that was about cyclic groups!

As for the function being a bijection:

We have $n$ lots of $gH$. We also have $n$ lots of $Hg^{-1}$ (proof as follows):

I really don't have any idea what can you possibly mean by "having lots of..."...

Since we have $n$ lots of $g$, we have at most $n$ lots of $Hg^{-1}$. We will only have less if, for some $k \in G$, $Hg^{-1}=Hk^{-1} \Rightarrow \ g^{-1}=k^{-1} \Rightarrow \ g=k$ (since inverses are unique).

This is incorrect: it can perfectly well be that $Ha^{-1}=Hb^{-1}$ without being true that $a=b$. You've to use the definition of the equivalence relation $Ha=Hb\,\,,\iff$...what?!

Therefore we have $n$ lots of $Hg^{-1}$.

Also, since inverses are unique, $g^{-1}$ is unique to a $g$. Therefore the function is injective.

We have a mapping from a set of $n$ elements to a set of $n$ elements which is injective. By the pigeonhole principle, the function is a bijection.
.

5. Do you mean $Ha=Hb \ \Leftrightarrow \ a=b$?

I don't think I know anything else connecting the two!

6. Originally Posted by Showcase_22
Do you mean $Ha=Hb \ \Leftrightarrow \ a=b$?

No! And if you don't remember this then no wonder you're having so much problem with this question. Perhaps it is a good idea to refresh this basic stuff now.

It is $Ha=Hb\Longleftrightarrow ab^{-1}\in H$ ...

Tonio

I don't think I know anything else connecting the two!
.

7. Thanks again Tonio. I have to admit that I have never seen that before. =S

However, doesn't that create a contradiction?

Suppose we have $g_1, \ g_2 \in G$ with $g_1 \neq g_2$.

Since $G$ is a group and is closed, $g_1^{-1}g_2 \in G$.

This implies that $Hg_1^{-1}=Hg_2^{-1}$.

This is also $f(g_1H)=f(g_2H)$ so the function isn't injective.

What's wrong with this argument?

(Also, sorry i'm being the bee in your bonnet! This question is really difficult for me).

8. Originally Posted by Showcase_22
Thanks again Tonio. I have to admit that I have never seen that before. =S

However, doesn't that create a contradiction?

Suppose we have $g_1, \ g_2 \in G$ with $g_1 \neq g_2$.

Since $G$ is a group and is closed, $g_1^{-1}g_2 \in G$.

This implies that $Hg_1^{-1}=Hg_2^{-1}$.

No! Please do pay attention: $G$ is a group, $H$ is a subgroup, then the equivalence relation is: for $g_1,g_2\in G\,,\,\,g_1\sim g_2\Longleftrightarrow g_1g_2^{-1}\in H$ ,and then we get (this requires a proof, of course) that $g_1g_2^{-1}\in H\Longleftrightarrow Hg_1=Hg_2$
This is VERY ,very basic stuff and you should know it and understand it before you attempt to solve your original problem.

Tonio

This is also $f(g_1H)=f(g_2H)$ so the function isn't injective.

What's wrong with this argument?

(Also, sorry i'm being the bee in your bonnet! This question is really difficult for me).
.

9. I haven't been getting anywhere with this function idea. I can't seem to prove anything with it!

I stumbled across this:

Suppose we have $n$ right cosets ( $|G:H|=n$, $G$ is a group and $H \subset G$).

Cosets are either equal or disjoint so they partition $G$ into a finite number of cosets (since $|G|$ is finite).

By Lagrange's theorem, $|H| | |G|$ ("order of H divides order of G", it looks confusing how i've written it!) so $|G|=n|H|$.

This proves the theorem (apparently).

I understand what it's done, but I fail to see how this actually works! Is it because we have $n|H|$ at the end and not $|H|n$?

10. Have you done Lagrange's theorem? It is easy to show that if $\left[G:H\right]_L$ and $\left[G:H\right]_R$ denote the number of left and right cosets respectively then $|H|\left[G:H\right]_R=|G|=|H|\left[G:H\right]_L\implies \left[G:H\right]_L=\left[G:H\right]_R$

11. We did it today! I decided to try using it for this question (since i've pretty much had it with that function!).

Was what i've written correct? I'm still trying to understand it properly. =S

Let $

\left[G:H\right]_R
$
be the number of right cosets of a group $G$. Then, since $H$ is a subgroup, $|H| | |G|$ or $|G|=n|H|$ since we know that $

\left[G:H\right]_R=n
$
.

$

Let \left[G:H\right]_L
$
be the number of left cosets of the same group. Then $|G|=k|H|$.

Then $n|H|=k|H|$ so $n=k$ so we have the same number of left and right cosets!!

12. Originally Posted by Showcase_22
We did it today! I decided to try using it for this question (since i've pretty much had it with that function!).

Was what i've written correct? I'm still trying to understand it properly. =S

Let $

\left[G:H\right]_R
$
be the number of right cosets of a group $G$. Then, since $H$ is a subgroup, $|H| | |G|$ or $|G|=n|H|$ since we know that $

\left[G:H\right]_R=n
$
.

$

Let \left[G:H\right]_L
$
be the number of left cosets of the same group. Then $|G|=k|H|$.

Then $n|H|=k|H|$ so $n=k$ so we have the same number of left and right cosets!!