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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 $Ha = Hb$


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.


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. – user17762 Oct 31 '11 at 21:52
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 $a=b$. – AMPerrine Oct 31 '11 at 21:53
@AMPerrine: So if $ah = bh'$, I need to show $ha = h'b$ for $h, h' \in H$? – Student Oct 31 '11 at 21:56
@Jon: Find a counterexample. – Mikko Korhonen Oct 31 '11 at 22:26
Note that for a counterexample you’ll need a non-Abelian group. What’s the simplest one that you know? – Brian M. Scott Oct 31 '11 at 22:47
up vote 2 down vote accepted

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 $b = \rho_{1}$

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

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


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

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

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. – Gerry Myerson Nov 1 '11 at 0:03
@Gerry: Thanks for pointing that out. – Student Nov 1 '11 at 0:13
Small note, a subgroup $H$ has the property "If $aH = bH$, then $Ha = Hb$" if and only if it is a normal subgroup. – Mikko Korhonen Nov 1 '11 at 9:01

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