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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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    @Soham You should add this all to your original question or ask another question by clicking the "$A$sk Questio$n$" link towards the top of the page. That way the question and answer will be searchable for anyone else who has questions about this proof.2012-07-29
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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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    @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."2012-07-29