Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


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

share|cite|improve this answer

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 $\,\phi(1)\ldots\,$ !

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.