TITLE: Left multiplication commuting functions (proof verification)
QUESTION [5 upvotes]: Let $G$ be a group and $g \in G$. Define $\lambda_g: G \rightarrow G$ by $\lambda_g(x) = gx$ for all $x \in G$. Similarly, define $\tau_g: G \rightarrow G$ by $\tau_g(x) = xg$ for all $x \in G$. If $\theta: G \rightarrow G$ is a bijection such that $\lambda_g \theta = \theta \lambda_g$ for all $g \in G$, show that $\theta = \tau_h$ for some $h \in G$.
My proof is as follows:
Since $\lambda_g \theta = \theta \lambda_g$ for all $g \in G$, we have
$$ g \theta(x) = \lambda_g(\theta(x)) = \theta (\lambda_g(x)) = \theta(gx) $$
for all $g, x \in G$. In particular, we have $$x^{-1} \theta(x) = \theta(x^{-1}x) = \theta(1)$$ or
$\theta(x) = x\theta(1)$ for all $x \in G$. Hence $\theta = \tau_{\theta(1)}$ as required.
It looks pretty simple, but I don't seem to use the condition that "$\theta$ is a bijection".
Is my proof correct? Thank you.

REPLY [4 votes]: The proof looks correct to me. As to the condition of $\theta$ being a bijection, rather than being an assumption this property is implied by the condition that $\lambda_g\theta=\theta\lambda_g$ for all $g\in G$.
Indeed, let $x\in G$. Setting $g=x(\theta(x))^{-1}$, we have
$$x = x(\theta(x))^{-1}\theta(x) = g\theta(x) = \lambda_g\theta(x) = \theta(\lambda_g(x)),$$
hence $x\in\theta(G)$. Thus, $\theta$ is surjective.
And from what you do, if $\theta(x)=\theta(y)$, then
$$\begin{align*}
\theta(1) &= \theta(x^{-1}x) = \theta\lambda_{x^{-1}}(x) = \lambda_{x^{-1}}\theta(x) = x^{-1}\theta(x)\\
\theta(1) &= \theta(y^{-1}y) = \theta\lambda_{y^{-1}}(y) = \lambda_{y^{-1}}\theta(y) = y^{-1}\theta(y) = y^{-1}\theta(x).
\end{align*}$$
Thus, $y^{-1}\theta(x)=x^{-1}\theta(x)$, so $y=x$. Thus, $\theta$ is also one-to-one.