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Prove that if $a$ is an element of a finite group $G$, then $ord(a)=ord(a^{-1})$.

So here's my proof, please let me know if I'm doing it right:

$ord(a)=n\implies a^n=e$
$e=a^n=a*a*a*...*a$
$e(a^{-1})^n=a*a*a*...*(a*a^{-1})*...*a^{-1}*a^{-1}*a^{-1}$
now the right side of the above equation is collapsing from the inside out
$(a^{-1})^n=e$
$\therefore ord(a)=n\implies a^n=e\implies (a^{-1})^n=e\implies ord(a^{-1})=n=ord(a)$

Does that look like a satisfactory proof, am I following the rules for groups?

Maths Rahul
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skwirlyburd
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    Strictly speaking, I think what you've shown is that the $|a^{-1}|\leq|a|$. Then, working in the opposite direction, you can show that $|a|\leq|a^{-1}|$. In addition, there is a hidden induction argument due to the $\dots$ (ellipses). – march Nov 09 '21 at 00:31
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    It is good but you are forgetting that $ord(a)$ is the smallest positive integer $n$ such that $a^n=e$. You proved only that $ord(a^{-1})\le n$. You can finish the proof by contradiction: if $(a^{-1})^k=e$, then $a^k=e$ by the same reasoning, so $k\ge n$. – Taladris Nov 09 '21 at 00:32
  • See also https://math.stackexchange.com/a/2794247/589 – lhf Nov 09 '21 at 00:34
  • From what I see you've only shown that $ord(a^{-1})\leq n$. I would consider $(a^{-1})^k=(a^{n-1})^k=a^{kn-k}$ and we have $a^q=e$ iff $q|n$ so you are searching for the smallest $k$ s.t $n|nk-k$ which is of course $k=n$ – includeCMath Nov 09 '21 at 00:34

2 Answers2

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Good work, you've got the main concept, but recall that the order of an element $a$ is the smallest positive integer such that $a^{ord(a)} = e.$ So for the proof to be complete you must also show that there is no smaller number $k$ such that $(a^{-1})^k = e.$

This is pretty easily done: suppose that there is some integer $k$ with $0 < k < n$ such that $(a^{-1})^k = e.$ Multiplying by $a^k$ on both sides we clearly get $e = a^k,$ but this implies that there is some $k$ with $a^k = e$ and $k < ord(a),$ which contradicts the definition of $ord(a).$ Therefore, no such $k$ can exist and $ord(a)$ is the smallest positive power of $a^{-1}$ which equals the identity element, so $ord(a^{-1}) = ord(a).$

Stephen Donovan
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It’s alright so far, but you still have not shown that $n$ is minimal. So strictly speaking you just have that the order of $a^{-1}$ is a divisor of the order of $a$. You can amend this quite easily though by using this result on $(a^{-1})^{-1}=a$.

A slightly different, interesting view on this: The set of powers $a^k$ forms a so called cyclic group, that is a group that is isomorphic to the additive group on $\mathbb Z/n\mathbb Z$ (where $n$ is the order of $n$). So the order of $a^k$ is the (additive) order of $k$ in $\mathbb Z/n\mathbb Z$. Thus if $k,n$ are coprime then $a^k$ does also have order $n$.

Lazy
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