# Proving two right cosets of $H$ in $G$ either are identical or have no element in common?

This question is a continuation from this question.

I.N.Herstein in Page 34 of his book "Topics in Algebra" defines a congruence relation like this:

DEFINITION: Let $G$ be a group , $H$ be 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 $ab^{-1} \in H$

Further he goes on to prove this Lemma 2.6 in page 35

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

PROOF: Let $[a] = \left \{x \in G |a \equiv x \mod H \right \}$. We first show that $Ha \subseteq [a]$. For if $h \in H$, then $a(ha)^{-1} = h^{-1}$ which is also in $H$. So by definition of congruence, this implies $ha \subseteq [a]$ for every $h \in H$ and so $Ha \subseteq [a]$.

My question:

To have $ab^{-1} \in H$ means $a,b \in G$ {since it is the necessity for congruence}. How does the proof go on to claim $Ha \subseteq [a]$ specifically? It should technically, claim $a,(ha) \in G$ which would imply $Ha \subseteq G$

Help much appreciated Soham

-

## 1 Answer

You want to show that $Ha \subseteq [a]$ yes? So take a general element say $ha$ on the left where $h \in H$. To prove that $ha \in [a]$ you need to prove that

$$ha \equiv a \mod H.$$

Now by definition $ha \equiv a \mod H$ if $a(ha)^{-1} \in H$. That is the definition of the congruence you gave above. However this is clear because $a(ha)^{-1} = h \in H$. It follows that $ha \equiv a \mod H$ so that $ha \in [a]$. It follows that

$$Ha \subseteq [a].$$

-
But of course! Thank you. You made the entire thing look so easy. :) Each time I missed one nuance of the proof or the other. Thank you. :) –  Soham Jul 30 '12 at 11:01