Problem

If \(x\) and \(g\) are elements of the group \(G\), prove that \(|x| = |g^{-1}xg|\). Deduce that \(|ab| = |ba|\) for all \(a, b \in G\).

Solution

Let \(d = |x|\). First, $$ (g^{-1}xg)^d = g^{-1}\underbrace{(xgg^{-1})…(xgg^{-1})}_{d-1\text{ times}}xg = g^{-1}x^{d-1}xg = g^{-1}x^dg = g^{-1}g = 1 $$

Next, suppose for a contradiction that there exists \(y < d\) such that \((g^{-1}xg)^y = 1\). Then, since \((g^{-1}xg)^y = g^{-1}x^yg\), it follows that

$$ g^{-1}x^yg = 1 $$

$$ g(g^{-1}x^yg)g^{-1} = g1g^{-1} $$

$$ x^y = 1 $$

This contradicts the fact that \(d\) is the smallest positive integer such that \(x^d = 1\).

Hence, \(d\) is the smallest positive integer that satisfies \((g^{-1}xg)^d = 1\), and by definition, \(d = |x| = |g^{-1}xg|\).

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

For any \(a,b \in G\), by replacing \(x\) with \(ab\) and \(g\) with \(a\) in the statement proven above,

$$ |ab| = |a^{-1}(ab)a| = |a^{-1}aba| = |ba| $$

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