Problem

Let \(G\) be a group and let \(G\) act on itself by left conjugation, so each \(g\in G\) maps \(G\) to \(G\) by $$ x \mapsto gxg^{-1}. $$

For fixed \(g\in G\), prove that conjugation by \(g\) is an isomorphism from \(G\) onto itself (i.e., is an automorphism of \(G\)). Deduce that \(x\) and \(gxg^{-1}\) have the same order for all \(x\) in \(G\) and that for any subset \(A\) of \(G\), \(|A| = |gAg^{-1}|\). Note here \(gAg^{-1} = \{gag^{-1} \mid a \in A\}\).

Solution

Let \(\varphi: G \rightarrow G\) be the mapping in discussion. That is, for fixed \(g\in G\) and any \(x \in G\), \(\varphi(x) = gxg^{-1}\).

For any \(x,y \in G\),

$$ \varphi(xy) = gxyg^{-1} = gx(g^{-1}g)yg^{-1} = (gxg^{-1})(gyg^{-1}) = \varphi(x)\varphi(y). $$

Therefore, \(\varphi\) is a homomorphism.

Define another mapping \(\psi: G \rightarrow G\) as \(\psi(x) = g^{-1}xg\). Then,

$$ \varphi\circ\psi(x) = \varphi(g^{-1}xg) = g(g^{-1}xg)g^{-1} = 1\cdot x\cdot 1 = x $$ and $$ \psi\circ\varphi(x) = \psi(gxg^{-1}) = g^{-1}(gxg^{-1})g = 1\cdot x\cdot 1 = x. $$

Therefore, \(\varphi\) and \(\psi\) are an inverse of each other, and thus, \(\varphi\) is a bijection.

Since \(\varphi\) is both a homomorphism and a bijection, by definition, it’s an isomorphism.

$$\tag*{$\blacksquare$}$$

Since \(\varphi: G \rightarrow G\) is a homomorphism, \(\varphi(1) = 1\). Let \(i\) be the order of \(x\). Then, using the fact that \(\varphi\) is a homomorphism,

$$ (gxg^{-1})^i = \varphi(x)^i = \varphi(x^i) = \varphi(1) = 1. $$

This shows that the order of \(gxg^{-1}\) is at most \(i\). Now let \(j\) be the order of \(gxg^{-1}\) and suppose \(j < i\) for a contradiction. Again, using the fact that \(\varphi\) is a homomorphism,

$$ 1 = (gxg^{-1})^j = \varphi(x)^j = \varphi(x^j). $$

Since the identity is unique in a group and \(\varphi(1) = 1\), \(x^j = 1\). This contradicts the fact that \(i\) is the order of \(x\) while \(j\) is smaller than \(i\). Therefore \(j\) must equal \(i\).

That is, \(x\) and \(gxg^{-1}\) have the same order for all \(x \in G\).

$$\tag*{$\blacksquare$}$$

\(gAg^{-1}\) is equivalent to \(\varphi(A)\). We have shown that \(\varphi\) is a bijection. Therefore,

$$ |A| = |gAg^{-1}|. $$

$$\tag*{$\blacksquare$}$$