# Understanding a proof about groups and cosets

1. Aug 31, 2013

### cragar

1. The problem statement, all variables and given/known data
H is a subgroup of G, and a and b are elements of G.

Show that Ha=Hb iff $ab^{-1} \in H$ .
3. The attempt at a solution

line 1: Then a=1a=hb for some h in H.
then we multiply both sides by b inverse.
and we get $ab^{-1}=h$
This is a proof in my book.
My question is on line 1 when they write 1a, is 1 the identity element in H.
So basically Ha=1a and this trick allows the rest to follow.

2. Aug 31, 2013

### CompuChip

H is a subgroup, so the identity in G is also the identity in H.

3. Sep 1, 2013

### CompuChip

Actually my answer was a bit short, so let me expand on it a bit more.

Recall the definition $Ha \equiv \{ ha \mid h \in H \}$. Let's consider the '=>' direction first and assume that $Ha = Hb$. Consider the fixed element $a$. Since $e \in H$ (I prefer e over 1 to indicate the identity element), $a = ea \in Ha$. On the other hand, since Ha = Hb, $a \in Hb$ meaning that there exists some $h \in H$ such that $a = hb$. From this is follows that $a b^{-1} (= h) \in H$ which is all you needed to show.

I assume that after that they proceed to show the converse, i.e. if $a b^{-1} \in H$ for the given elements $a, b$ then $Ha = Hb$. I imagine that proof will run along the lines of: assume $ab^{-1} \in H$, let $h \in H$ be arbitrary, then $h a = h a (b^{-1} b) = h (a b^{-1}) b$ and since H is a subgroup, $h (ab^{-1})$ is the product of two elements in H which is itself in H, so $h a = h' b$ with $h' ( = h a b^{-1}) \in H$, hence $Ha \subseteq Hb$ - the proof for $Hb \subseteq Ha$ follows similarly.

Note that actually $\forall a, b \in G \left( a, b \in H \implies ab^{-1} \in H \right)$ is a necessary and sufficient condition for H to be a subgroup of G. So basically your theorem is "Ha = Hb iff H is a subgroup" or "For subgroups, and subgroups only, all right cosets are equal".

Last edited: Sep 1, 2013
4. Sep 13, 2013

ok thanks