Exercise 1.1.22 ($|ab| = |ba|$)

Question

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

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
Since $(g^{-1}xg)^n = g^{-1} x^n g$ for all $n \in \Z$, 


$$x^n = 1 \iff g^{-1} x^n g = 1 \iff (g^{-1}x g)^n = 1,$$
so $x$ has finite order if and only if $g^{-1}x g$ has finite order.

\bigskip
If $x$ has finite order,
$$n \mid |x| \iff x^n = 1 \iff g^{-1} x^n g = 1 \iff (g^{-1}x g)^n = 1 \iff n \mid |g^{-1}xg|.$$
Therefore, in both cases
$$|x| = |g^{-1}x g|.$$
Since 
$$ab = a(ba) a^{-1} = g^{-1} (ba) g \qquad (	ext{where }g =a^{-1} \in G),$$
then
$$|ab | = |g (ab) g^{-1}| = |ba|.$$
\end{proof}

Exercise 1.1.22 ($|ab| = |ba|$)

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Exercise 1.1.22 ($|ab| = |ba|$)

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