Exercise 4.3.36 (Left and right regular representations)

Question

Let $\pi : G 	o S_G$ be the left regular representation afforded by the action of $G$ on itself by left multiplication. For each $g \in G$ denote the permutation $\pi(g)$ by $\sigma_g$, so that $\sigma_g(x) = gx$ for all $x \in G$. Let $\lambda : G 	o S_G$ be the permutation representation afforded by the corresponding right action of $G$ on itself, and for each $h \in G$ denote the permutation $\lambda(h)$ by $	au_h$. Thus $	au_h(x) = xh^{-1}$ for all $x \in G$ ($\lambda$ is called the {\bf right regular representation of G}).
\begin{enumerate}
\item[{\bf(a)}] Prove that $\sigma_g$ and $	au_h$ commute for all $g,h \in G$. (Thus the centralizer in $S_G$ of $\pi(G)$ contains the subgroup $\lambda(G)$, which is isomorphic to $G$).
\item[{\bf(b)}] Prove that $\sigma_g = 	au_g$ if and only if $g$ is an element of order $1$ or $2$ in the center of $G$.
\item[{\bf(c)}] Prove that $\sigma_g = 	au_h$ if and only if $g$ and $h$ lie in the center of $G$ (*). Deduce that ${\pi(G) \cap \lambda(G) = \pi(Z(G)) = \lambda(Z(G))}$.
\end{enumerate}

\bigskip
(*) We must add `` ... and $g = h^{-1}$'' . See the note after the proof (note of R.G.)

Richard Ganaye · Accepted Answer

\begin{proof} Let $\pi : G 	o S_G$ be the left regular representation, and $\lambda: G 	o S_G$ be the right regular representation. Then $\sigma_g$ and $	au_h$ are defined for all $g,h \in G$ by
$$
\pi(g) = \sigma_g
\left\{
\begin{array}{ccl}
G &	o & G\
x & \mapsto & gx
\end{array}
ight.
,
\qquad
\lambda(h) = 	au_h
\left\{
\begin{array}{ccl}
G &	o & G\
x & \mapsto & xh^{-1}
\end{array}
ight.
.
$$
\begin{enumerate}
\item[{\bf(a)}] Then for all $x \in G$,
\begin{align*}
(\sigma_g \circ 	au_h)(x) &= \sigma_g(xh^{-1}) = gxh^{-1},\
(	au_h \circ \sigma_g)(x) &= 	au_h(gx) = gxh^{-1}.
\end{align*}
Therefore $\sigma_g \circ 	au_h = 	au_h \circ \sigma_g$, so  $\sigma_g$ and $	au_h$ commute for all $g,h \in G$.
\item[{\bf(b)}] Suppose that $\sigma_g = 	au_g$. Then for all $x \in G$, $gx = xg^{-1}$. In particular, for $x = 1$, $g = g^-1$, so $g^2 = 1$. This shows that $g$ is an element of order $1$ or $2$. Moreover, since $g = g^{-1}$, $gx= xg^{-1} = xg$, for every $x \in G$, thus $g \in Z(G)$.

Conversely, suppose that $g^2 = 1$ and $g \in Z(G)$. Then $g = g^{-1}$ and for all $x \in G$,
$$\sigma_g(x) = g x = x g = xg^{-1} =	au_g(x),$$
so $\sigma_g = 	au_g$.

 In conclusion, $\sigma_g = 	au_g$ if and only if $g$ is an element of order $1$ or $2$ in the center of $G$.
 
 
\item[{\bf(c)}] Suppose that $\sigma_g = 	au_h$. Then for all $x \in G$, $gx = xh^{-1}$. In particular, for $x = 1$, we obtain $g = h^{-1}$. Therefore $g x = x g$ for all $x \in G$, so $g \in Z(G)$, and $h = g^{-1} \in Z(G)$.

Conversely, if $g,h \in Z(G)$ and $g = h^{-1}$, then for all $x \in G$, $gx = xg$ and $hx = xh$. Then
$$\sigma_g(x) = gx = xg = xh^{-1} = 	au_h(x),$$
so $\sigma_g = 	au_h$.

In conclusion $\sigma_g = 	au_h$ if and only if $g,h \in Z(G)$ and $g= h^{-1}$. (*) 

Note first that $\pi(Z(G)) = \lambda(Z(G))$: if $f \in \pi(Z(G))$, then $f = \sigma_g$ for some $g \in Z(G)$. For all $x \in G$,
\begin{align*}
\sigma_g(x) &= gx\
&= xg \
&= 	au_{g^{-1}}(x).
\end{align*}
Therefore $f = \sigma_g = 	au_{g^{-1}}$, where $g^{-1} \in Z(G)$, thus $f \in \lambda(Z(G))$. This proves $\pi(Z(G)) \subseteq \lambda(Z(G))$, and similarly $ \lambda(Z(G)) \subseteq \pi(Z(G))$, so
\begin{align}
\pi(Z(G)) = \lambda(Z(G)).
\end{align}
Now, if $ f \in  \pi(Z(G))\cap   \lambda(Z(G))$, then $f = \sigma_g = 	au_h$ for some elements $g,h \in G$. We have proved that in this case $g \in Z(G)$, thus $f \in \pi(Z(G))$.

Conversely, if $f \in \pi(Z(G))$, then $f \in \pi(Z(G))\cap   \lambda(Z(G))$ by (1), and $\pi(Z(G))\cap   \lambda(Z(G)) \subseteq \pi(G) \cap \lambda(G)$, thus $f \in \pi(G) \cap \lambda(G)$.

This shows that
$$ \pi(Z(G))\cap   \lambda(Z(G)) = \pi(Z(G)) = \lambda(Z(G)).$$


\end{enumerate}
\end{proof}

(*) Note : The proposition given in the statement is false in its original form. As a counterexample, consider $G = D_8$. Then $g = 1$ and $h = r^2$ are in the center of $G$, but $\sigma_g = \mathrm{id}_G$, and $	au_h(1)  = h^{-1} = r^2 
e 1$, so $	au_h 
e  \mathrm{id}_G = \sigma_g$.

Exercise 4.3.36 (Left and right regular representations)

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Exercise 4.3.36 (Left and right regular representations)

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