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Let $G$ be any group, let $r_x,l_x: G \to G$ defined by $r_x(g)=gx$, and $l_x(g)=xg$. Let $R=\{r_x:x \in G\}$ and $L=\{l_x:x \in G\}$. Show that

$L=\{f \in \mathrm{Sym}(G): \forall r \in R~~~ fr=rf\}$

I have thought about it for hours, but I can't prove the right side belongs to the left side. Please help me.

2 Answers 2

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The right side consists of the functions $f$ that are permutations of $G$ such that for all $r_x \in R$ and $g \in G$ we have $f(gx) = f(g)x .$ In particular, for $g=e$ we have $ f(x)= f(e) x $. So $f = l_{f(e)}.$

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    Thank you, it's my first question here, luckily it's solved so quickly.2012-04-21
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The inclusion that you have yet to prove is obtained as follows:

  1. The condition $f\circ r=r\circ f,\ \forall r\in R$ explicitly means that $f(gx)=f(g)x,\quad\forall g,x\in G.$

  2. If we fix $g$ equal to the neutral element $e$ of $G$ in the previous condition, then we get $f(x)=f(e)x,\quad\forall x\in G.$

This is what you want, i.e. $f=l_{f(e)}\in L.$