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Let $H$ be a subgroup of a group $G$ and let $a, b \in G$. I need to give a counterexample or proof of the following statement:

If $aH = bH$, then $Ha = Hb$

Proof:

For every $h \in H, ah = bh$

$ \begin{align} ah &= bh \newline ahh^{-1} &= bhh^{-1} \newline a &= b \newline ha &= hb \end{align} $

Could someone critic my proof?

Thanks in advance.


Edit

Look at the answer below.

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    Looks good; why don’t you go ahead and write it up as an answer.2011-10-31

1 Answers 1

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Counterexample: We want to use a non-Abelian group such as $S_{3}$

$\mu_{1} = \pmatrix{1&2&3\\1 &3 &2}$, $\mu_{2} = \pmatrix{1&2&3\\3 &2 &1}$, $\mu_{3} = \pmatrix{1&2&3\\2 &1 &3}$

$\rho_{0} = \pmatrix{1&2&3\\1 &2 &3}$, $\rho_{1} = \pmatrix{1&2&3\\2 &3 &1}$, $\rho_{2} = \pmatrix{1&2&3\\3 &1 &2}$

Let $H = \{\rho_{0}, \mu_{2}\}$, $a = \mu_{1}$ and $b = \rho_{1}$

$aH = \mu_{1}\{\rho_{0}, \mu_{2}\} = \{\mu_{1}, \rho_{1}\}$

$bH = \rho_{1} \{\rho_{0}, \mu_{2}\} = \{\rho_{1}, \mu_{1}\}$

But

$Ha = \{\rho_{0}, \mu_{2}\}\mu_{1} = \{\mu_{1}, \rho_{2}\}$

$Hb = \{\rho_{0}, \mu_{2}\}\rho_{1} = \{\rho_{1}, \mu_{3}\}$

So $Ha \neq Hb$

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    Small note, a subgroup $H$ has the property "If $aH = bH$, then $Ha = Hb$" if and only if it is a normal subgroup.2011-11-01