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I think one way to solve it is by considering the conjugation of $a$ and $a^{-1}$ using $g \in G$. That is, $a^g=gag^{-1}$ and $(a^{-1})^g= ga^{-1}g^{-1}$. Then, if $n$ is the order of $a$ we have $(a^g)^n=e$; since,

$(a^g)^n=gag^{-1} gag^{-1} \cdots g^{-1} gag^{-1} =$ $= gaegaea \cdots ag^{-1} =ga^ng^{-1}=geg^{-1}=e$

and $((a^{-1})^g)^n=e$; since,

$((a^{-1})^g)^n=ga^{-1}g^{-1}ga^{-1}g^{-1} \cdots g^{-1}ga^{-1}g^{-1}= ga^{-1}ea^{-1}e\cdots a^{-1}g^{-1}= g(a^{n})^{-1}=geg^{-1}=e$

Therefore, $|a|=|a^{-1} |$. I am right? If not, how else can I do it?

Hopmaths
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2 Answers2

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Suppose $a\neq e$ (the result is obvious if $a=e$)

$aa^{-1}=a^{-1}a=e$ implies that $e=(aa^{-1})^=a^n(a^{-1})^n$.

Let $N(a)$ the set of positive integers such that $a^n=e$.

Suppose that $a^n=e$, we deduce that $a^n(a^{-1})^n=(a^{-1})^n=e$.

Suppose that $(a^{-1})^n=e, a^n(a^{-1})^n=a^n=e$ we deduce that $N(a)=N(a^{-1})$, since $ord(a)$ is the smallest element of $N(a)$, $a$ and $a^{-1}$ have the same order.

Tsemo Aristide
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We can prove by contradiction for example. Let $n$ be order of $a$ and $m$ be the order of $a^{-1}$, wlog suppose $n>m$, then $e=a^n=a^n\cdot e=a^n \cdot \left(a^{-1}\right)^m=a^{n-m}$, but $n-m<n$ so this can't be by the definition of order as $a$ has order $n$. Contradiction.

Alexey Burdin
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