# Left and Right cosets.

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• Jan 30th 2010, 10:08 AM
Showcase_22
Left and Right cosets.
Quote:

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 $\displaystyle |G:H|=|G|/|H|=n$ for $\displaystyle n \in \mathbb{N}$.

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

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

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

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

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

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

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

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

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?
• Jan 30th 2010, 10:27 AM
tonio
Quote:

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

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

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

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

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

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

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

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

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

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

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 $\displaystyle X_L\,,\,\,X_R$ represent the sets of left and right cosets of $\displaystyle G\,\,\,in\,\,\,H$ , and define

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

Tonio
• Jan 30th 2010, 11:25 AM
Showcase_22
I got caught up on the question before that was about cyclic groups!

As for the function being a bijection:

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

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

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

We have a mapping from a set of $\displaystyle n$ elements to a set of $\displaystyle n$ elements which is injective. By the pigeonhole principle, the function is a bijection.
• Jan 30th 2010, 01:15 PM
tonio
Quote:

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

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

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

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

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

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

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

.
• Jan 30th 2010, 08:45 PM
Showcase_22
Do you mean $\displaystyle Ha=Hb \ \Leftrightarrow \ a=b$?

I don't think I know anything else connecting the two!
• Jan 31st 2010, 12:12 AM
tonio
Quote:

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

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

Tonio

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

.
• Jan 31st 2010, 12:32 PM
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 $\displaystyle g_1, \ g_2 \in G$ with $\displaystyle g_1 \neq g_2$.

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

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

This is also $\displaystyle f(g_1H)=f(g_2H)$ 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).
• Jan 31st 2010, 06:16 PM
tonio
Quote:

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 $\displaystyle g_1, \ g_2 \in G$ with $\displaystyle g_1 \neq g_2$.

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

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

No! Please do pay attention: $\displaystyle G$ is a group, $\displaystyle H$ is a subgroup, then the equivalence relation is: for $\displaystyle g_1,g_2\in G\,,\,\,g_1\sim g_2\Longleftrightarrow g_1g_2^{-1}\in H$ ,and then we get (this requires a proof, of course) that $\displaystyle g_1g_2^{-1}\in H\Longleftrightarrow Hg_1=Hg_2$
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 $\displaystyle f(g_1H)=f(g_2H)$ 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).

.
• Feb 2nd 2010, 01:09 PM
Showcase_22
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 $\displaystyle n$ right cosets ($\displaystyle |G:H|=n$, $\displaystyle G$ is a group and $\displaystyle H \subset G$).

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

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

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 $\displaystyle n|H|$ at the end and not $\displaystyle |H|n$?
• Feb 2nd 2010, 01:13 PM
Drexel28
Have you done Lagrange's theorem? It is easy to show that if $\displaystyle \left[G:H\right]_L$ and $\displaystyle \left[G:H\right]_R$ denote the number of left and right cosets respectively then $\displaystyle |H|\left[G:H\right]_R=|G|=|H|\left[G:H\right]_L\implies \left[G:H\right]_L=\left[G:H\right]_R$
• Feb 2nd 2010, 01:25 PM
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 $\displaystyle \left[G:H\right]_R$ be the number of right cosets of a group $\displaystyle G$. Then, since $\displaystyle H$ is a subgroup, $\displaystyle |H| | |G|$ or $\displaystyle |G|=n|H|$ since we know that $\displaystyle \left[G:H\right]_R=n$.

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

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

Please say that's right!!
• Feb 2nd 2010, 01:34 PM
Drexel28
Quote:

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

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

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

Please say that's right!!

Strange way of phrasing it, but yes.
• Feb 2nd 2010, 01:35 PM
Showcase_22
Thanks Drexel! (for helping me with this, and the other question I posted a while ago).