TITLE: If $G$ is a finite group of order $n$, why is it isomorphic to its centralizer in $S_n$?
QUESTION [5 upvotes]: If $G$ is a finite group of order $n$, why is it isomorphic to its centralizer in $S_n$? Here, we embed $G$ in $S_n$ via the left regular representation.
From thinking a bit about the classification of covering spaces in algebraic topology, I suspect that this must be true, and I expect that I can probably prove it using topology if I wanted to.
Is there a representation/group-theoretic way of proving this?
Also, is the converse true? Ie, if $G$ is a subgroup of $S_n$ that is isomorphic to its own centralizer, then it must be transitive and have order $n$?

REPLY [2 votes]: Suppose that $f: G \rightarrow G$ is a bijection that commutes with every left translation. Then $f(xy) = x f(y)$ for all $x, y \in G$. In particular choosing $y = 1$ gives $f(x) = xf(1)$ for all $x \in G$. Thus $f$ is a right translation by $f(1)$. Also, any right translation commutes with a left translation, so the centralizer of the left regular representation in $S_n$ is the right regular representation. Similarly, the centralizer of the right regular representation in $S_n$ is the left regular representation.