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Given a group $G$, let $L$ and $R$ be the subgroups of $S=\operatorname{Sym}(G)$ consisting of all left multiplications by elements of $G$ and all right multiplications by elements of $G$, respectively. Show that $L=C_S(R)$ and $R=C_S(L)$ (the centralizers of $R$ and $L$, respectively).

It's easy to show that $L \subseteq C_S(R)$:

Let $\phi_g \in L$,$\,\,\psi_h \in R$, and $x \in G$ ($\phi_g(x)=gx$, and similar for $\psi_h$). Then $(\phi_g \circ \psi_h)(x)=\phi_g(xh)=gxh=\psi_h(gx)=\psi_h(\phi_g(x))=(\psi_h \circ \phi_g)(x)$. It follows that $L \subseteq C_S(R)$.

However, I'm having a hard time showing that $C_S(R) \subseteq L$. What I've done is this:

Let $\phi \in C_S(R)$ and let $\psi_h \in R$ and $x \in G$. Then we have $(\phi \circ \psi_h)(x)=\phi(\psi_h(x))=\phi(xh)=\psi_h(\phi(x))=(\psi_h \circ \phi)(x)$.

The result is so intuitively clear, but I don't know how to show that $\phi$ must necessarily be given by a left multiplication. How do we go about showing this?


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Hint: consider $x=1$ in your equation. The element you want would be... –  awllower Mar 22 '13 at 15:20

2 Answers 2

up vote 3 down vote accepted

Consider subsititute your equation by $x=1$.

Answer: We shall have: $(\phi \circ \psi_h)(1)=\phi(\psi_h(1))=\phi(h)=\psi_h(\phi(1))=\phi (1)h$. And so $\phi$ is given by left multiplication by $\phi(1)$.

Inform me of any disdains. Thanks very much.

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An idea; it's the same to take left mult. by the left with elements or with their inverses, so:

$$\forall\;\phi\in C_S(R)\;,\;\psi_{h^{-1}}\in R\;,\;x\in G:$$


But choosing $\,h:=x\,$ we get $\,\phi(1)=\phi(x)x^{-1}\iff\phi(x)=\phi(1)x\Longrightarrow \,\,\phi$ indeed is a left multiplication: by $\,\phi(1)\ldots\,$ !

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