Cosets

**matt.qmar** · November 6th 2010, 01:50 PM

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 $\in$ G, and aH is the coset of H containing a, defined as $aH = \{ah\ : h \in H\}$

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

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

I'm not sure where to start on these.

Thanks!

**h2osprey** · November 6th 2010, 03:43 PM

1.

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

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

2.

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

**matt.qmar** · November 6th 2010, 04:26 PM

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

Thanks!!

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