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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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    $aH = bH$ means for any $h_1 \in H$, there exists $h_2 \in H$ such that $ah_1 = bh_2$ and for any $h_3 \in H$, there exists $h_4 \in H$ such that $bh_3 = ah_4$. It doesn't mean $ah = bh$ for all $h \in H$. The equality holds only as sets.2011-10-31
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    You can't assume that "For every $h\in H,ah=bh$," but rather that $ah$ is equal to some $bh'\in bH$. You can tell there's been a problem when you wind up showing that $a=b$.2011-10-31
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    @AMPerrine: So if $ah = bh'$, I need to show $ha = h'b$ for $h, h' \in H$?2011-10-31
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    That's for any $h\in H$, there exists an $h'\in H$ such that $ha=h'b$. Just be aware that the $h'$ in the left coset equation need not be the same as the $h'$ for the right coset (so you might want to denote it differently for clarity's sake). And as Sivaram Ambikasaran pointed out above, you must prove inclusion in both directions.2011-10-31
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    @AMPerrine: Could I get a hint for this problem? I can't seem to make any progress at all.2011-10-31
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    @Jon: Find a counterexample.2011-10-31
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    @m. k.: Good point. I think I lost sight of the original problem somewhere along the way.2011-10-31
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    Note that for a counterexample you’ll need a non-Abelian group. What’s the simplest one that you know?2011-10-31
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    @Brian: Symmetries on an equilateral triangle $S_{3}$2011-10-31
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    I thought that would probably be it. Good: it’s a good place to start looking for a counterexample.2011-10-31
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    @Brian: Could you take a look at my edit? Thanks2011-10-31
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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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    You are assuming, contrary to fact, that everyone uses the same notation for the elements of $S_3$. Please include in your answer what you mean by $\rho_0$, $\mu_2$, etc., as without that information it is impossible to evaluate your answer.2011-11-01
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    @Gerry: Thanks for pointing that out.2011-11-01
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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