Prove that in any group an element and its inverse have same order.

Question

Prove that an element and its inverse have same order in any group.

Answer 1

$$(g^{-1})^{o(g)}=(g^{o(g)})^{-1}=e^{-1}=e$$ proves that $o(g^{-1})\leq o(g)$.

$$g^{o(g^{-1})}=((g^{-1})^{-1})^{o(g^{-1})}=((g^{-1})^{o(g^{-1})})^{-1}=e^{-1}=e$$ proves that $o(g)\leq o(g^{-1})$

Answer 2

Since $g^{-1}\in \langle g\rangle$, we have that $$\langle g^{-1}\rangle\subseteq\langle g\rangle$$ For the same reason $$\langle g\rangle\subseteq\langle g^{-1}\rangle$$

The order of an element $g$ is infinite if $\langle g\rangle$ is infinite, otherwise it's the same as $|\langle g\rangle|$ (number of elements).