# Right Coset and the associated lemma from Herstein: Appreciating whats going on?

I.N. Herstein in Page 34 (last line) and Page 35 of "Topics in Algebra" book goes on to explain a definition of right coset and a lemma like this:

Def: If $H$ is a subgroup of G, and $a \in G$, then $Ha = \left \{ha|h\in H \right \}$;then $Ha$ is the right coset of $H$ in $G$

Lemma: FOr all $a \in G$ $Ha = \left \{x \in G |a \equiv x mod H \right \}$

He goes on to define a set $[a]$ exactly like $Ha$ and trying to show $Ha \subseteq [a]$

My confusion:

1. Whats going on here?

2. More specifically, what the lemma trying to convey and why did the author go on to define $[a]$ exactly like $Ha$ and trying to show $Ha \subseteq [a]$ Isnt it trivial that every set is a subset of itself?

3. If you have the proof of the lemma with you, can you help me understand it. I am not able to understand why exactly are we dealing with $a(ha)^{-1}$ which I understand as motivated from $a = ha mod H$

Thanks for your time and patience Soham

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The bracket notation is usually used in reference to equivalence relations. If $\sim$ is an equivalence relation on a set $G$, then $[a] = \{x \in G : x \sim a\}$.

So if I had to guess what is going on here, I would say that he defines $Ha = \{ha : h \in H\}$. Then he would have defined an equivalence relation $\sim$ on $G$ by $a \sim b$ if and only if $ab^{-1} \in H$.

Then it is easy (but not trivial) to show that $Ha \subset [a]$.

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He goes on to show that $Ha \equiv [a]$ How exactly does he prove it, it would be helpful to know –  Soham Jul 29 '12 at 21:58
@Soham Suppose $b \in Ha$, then $b = ha$. $ba^{-1} = h$. $ab^{-1} = h^{-1} \in H$. So $a \simeq b$. $b \in [a]$. Hence $Ha \subset [a]$. Suppose $b \in [a]$. So $a \simeq b$. So $ab^{-1} \in H$. So $ab^{-1} = h$. $b^{-1} = a^{-1}h$ so $b = ha$. Hence $b \in Ha$. So $[a] \subset Ha$. –  William Jul 29 '12 at 22:02
I understand Herstein used the definition of congruence to prove that $b \in [a]$. I.e if $a \equiv b mod H$ and it can be shown $a,b \in G$ and $H$ is a subgroup of $G$ then, $ab^{-1} \in H$ However how does he claim $ha \in [a]$ at best he can claim $ha \in G$ –  Soham Jul 29 '12 at 22:13
My definition of $a \sim b$ is $a \equiv b \text{ mod } H$. For your second qustion, $ha \in [a]$ if and only if $ha \sim a$ if and only if $haa^{-1} \in H$ if and only if $h \in H$, which is true. –  William Jul 29 '12 at 22:24
Herstein defines congreuence as : DEFINITION. Let G be a group, H a subgroup of G; for $a, b \in G$ we say a is congruent to b mod H, written as $a \equiv b mod H$ if $ab^{- 1} \in H$. My comments So here, $a,b$ have to be the element of the group isnt it, and not of the subgroup $H$, so if we can prove $ab^{-1} \in H$ the job is done as far to prove $a,b \in G$ –  Soham Jul 29 '12 at 22:31

The interesting thing about cosets is that they are equivalence classes in the group $G$. This means that either $Ha \cap Hb = \emptyset$ or $Ha = Hb$ for any $a, b \in G$. This is a very important fact which will be used to prove Lagrange's Theorem.

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Thanks I did a bit more reading in the meanwhile and I did get this. Can you help me understand the proof? –  Soham Jul 29 '12 at 22:05
@Soham Someone here will certainly be able to answer your questions. Please edit your question or post a new one giving the full lemma and proof. Also include some definitions to make sure we all understand the terminology and notation used in your textbook. For example, I am unfamiliar with the notation $a \equiv x \mod H$. You should also try as best you can to ask specific questions about the parts in the proof that you don't understand. –  Code-Guru Jul 29 '12 at 22:21
Hi Thanks. How can I close this question or retire it. I will try to form a completely new question, as I have certainly gained some insight from Williams answer and a little bit of introspection from my side. –  Soham Jul 29 '12 at 22:23
@Soham You should click the check mark to accept an answer. Other answers may still be posted if anyone wants to take the time. You should also vote for any answers that you find helpful, even if you can only chose one as "the accepted answer." –  Code-Guru Jul 29 '12 at 22:25