## Interaction of the left and right regular representations of a group

Let $\pi : G \rightarrow S_G$ be the left regular representation of $G$ and let $\lambda : G \rightarrow S_G$ be the right regular representation. That is, for all $g, x \in G$, $\pi(g)(x) = gx$ and $\lambda(g)(x) = xg^{-1}$.

1. Prove that $\pi(g)$ and $\lambda(h)$ commute for all $g,h \in G$. (Thus $\pi[G] \leq C_{S_G}(\lambda[G])$ and $\lambda[G] \leq C_{S_G}(\pi[G])$.)
2. Prove that $\pi(g) = \lambda(g)$ if and only if $g \in G$ is an element of order 1 or 2 and $g \in Z(G)$.
3. Prove that $\pi(g) = \lambda(h)$ if and only if $g,h \in Z(G)$ and $g = h^{-1}$. Deduce that $\pi[Z(G)] = \pi[G] \cap \lambda[G] = \lambda[Z(G)]$.

1. Let $g,h,x \in G$. Now $[\pi(g) \circ \lambda(h)](x) = \pi(g)(\lambda(h)(x))$ $= \pi(g)(xh^{-1})$ $= g(xh^{-1})$ $= (gx)h^{-1}$ $= \lambda(h)(gx)$ $= \lambda(h)(\pi(g)(x))$ $= [\lambda(h) \circ \pi(g)](x)$. Thus $\pi(g) \circ \lambda(h) = \lambda(h) \circ \pi(g)$ for all $g,h \in G$. In other words, $\pi[G] \leq C_{S_G}(\lambda[G])$ and $\lambda[G] \leq C_{S_G}(\pi[G])$.
2. $(\Rightarrow)$ Suppose $\pi(g) = \lambda(g)$. Then for all $x \in G$, $gx = xg^{-1}$. Letting $x = 1$ we see that $g = g^{-1}$; thus either $g = 1$ or $|g| = 2$. Moreover, we have $gx = xg$ for all $x \in G$; thus $g \in Z(G)$.

$(\Leftarrow)$ Suppose $g \in Z(G)$ is either 1 or has order 2. Then $g = g^{-1}$. Then we have $\pi(g)(x) = gx = xg = xg^{-1} = \lambda(g)(x)$ for all $x \in G$; hence $\pi(g) = \lambda(g)$.

3. $(\Rightarrow)$ Suppose $\pi(g) = \lambda(h)$. Then for all $x \in G$, $gx = xh^{-1}$. For $x = 1$ this gives $g = h^{-1}$. Thus we have $gx = xg$ and $hx = xh$ for all $x \in G$; hence $g,h \in Z(G)$.

$(\Leftarrow)$ Suppose $g,h \in Z(G)$ and $g = h^{-1}$. Then for all $x \in G$, we have $\pi(g)(x) = gx = xg = xh^{-1} = \lambda(h)(x)$; thus $\pi(g) = \lambda(h)$.

Now if $\sigma \in \pi[G] \cap \lambda[G]$, then $\pi(g) = \sigma = \lambda(h)$ for some $g, h \in G$. As we saw above, $g, h \in Z(G)$. Thus $\pi[G] \cap \lambda[G]$ is contained in both $\pi[Z(G)]$ and $\lambda[Z(G)]$.

Suppose $\sigma \in \pi[Z(G)]$. Then $\sigma = \pi(g)$ for some $g \in Z(G)$. Now $\sigma(x) = \pi(g)(x) = gx = xg = \lambda(g^{-1})(x)$, hence $\sigma \in \pi[G] \cap \lambda[G]$. Similarly, $\lambda[Z(G)] \subseteq \pi[G] \cap \lambda[G]$.

In my book, $g=h^{-1}$ is not in 3. Is it an error in the book?
I haven’t checked myself, but I imagine that the statement is false if we drop $g=h^{-1}$.