Question

If $\mathit{a}\mathbf{\in }\mathit{G}$,prove that $\left|a\right|\mathbf{=}\left|a^{-1}\right|$.

Short answer

It is proved that$\left|a\right|=\left|a^{-1}\right|$.

Step-by-step solution

Write the properties ofG

If$G$is a group and $a\in G$an element of finite order $n$then $a^k=e$ if and only if $n\left|k\right.$.

Show that |a|=|a-1|

Prove that if$a^k=e$ then, ${\left(a^{-1}\right)}^k=e$.

Assume $a^k=e$ for some integer $k$.

Let ${\left(a^{-1}\right)}^k=h\in G$.

Multiply by $a^k$ from the left then we have,

$e=h{a}^k=he=h$

Conversely,

Assume${\left(a^{-1}\right)}^k=e$for some integer $k$.

Assume $a^k=h\in G$then,

$e={\left(a^{-1}\right)}^{k}h=h$

Hence Proved.