1. ## Cosets

Hello,

I am trying to prove, or disprove by means of a counterexample the two statements.

For a subgroup H of a group G, with elements a,b $\displaystyle \in$ G, and aH is the coset of H containing a, defined as $\displaystyle aH = \{ah\ : h \in H\}$

(1)
$\displaystyle aH=bH$ implies $\displaystyle Ha^{-1} = Hb^{-1}$

(2)
$\displaystyle aH = bH$ implies $\displaystyle Ha = Hb$

I'm not sure where to start on these.

Thanks!

2. 1.

$\displaystyle aH = bH \iff \{ah | h \in H\} = \{bh | h \in H\} \iff \{(ah)^{-1} | h \in H\} = \{(bh)^{-1} |h \in H\}$

And you should be able to see with further manipulations that the implication is indeed true.

2.

Look at the group $\displaystyle S_3$. Can you find 2 elements that show this implication is not true?

3. Just to confirm, in (1), the next step would be
$\displaystyle \{(ah)^{-1} | h \in H\} = \{(bh)^{-1} |h \in H\}$
then
$\displaystyle \{(h)^{-1}(a)^{-1} | h \in H\} = \{(h)^{-1}(b)^{-1} |h \in H\}$
by properties of inverse, then
$\displaystyle \{(h)(a)^{-1} | h \in H\} = \{(h)(b)^{-1} |h \in H\}$
Since if $\displaystyle h \in H, \rightarrow h^{-1} \in H$
Then we have the desired result!

Thanks!!