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If $a$ is a group element, prove that $a$ and $a^{-1}$ have the same order.

I tried doing this by contradiction.

Assume $|a|\neq|a^{-1}|$.

Let $a^n=e$ for some $n\in \mathbb{Z}$ and $(a^{-1})^m=e$ for some $m\in \mathbb{Z}$, and we can assume that $m < n$.

Then $e= e*e = (a^n)((a^{-1})^m) = a^{n-m}$. However, $a^{n-m}=e$ implies that $n$ is not the order of $a$, which is a contradiction and $n=m$.

But I realized this doesn’t satisfy the condition if $a$ has infinite order. How do I prove that piece?

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    $\begingroup$ Note that this is exercise 4 in chapter 3 in Gallian's Contemporary Abstract Algebra. $\endgroup$
    – a student
    Dec 22, 2015 at 11:05
  • $\begingroup$ If you know that the order of an element equals the order of the subgroup generated by it, you just have to know that an element generates the same subgroup as its inverse (as it and its inverse are contained in the subgroup generated by the other). $\endgroup$
    – j.p.
    Nov 9, 2021 at 6:59
  • $\begingroup$ See also math.stackexchange.com/questions/2794098/… $\endgroup$
    – lhf
    Nov 10, 2021 at 10:23

4 Answers 4

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Let $a^n$ be $e$, then $e=(aa^{-1})^n=a^n(a^{-1})^n=e(a^{-1})^n=(a^{-1})^n$.

Let $(a^{-1})^n=e$, then $e=(aa^{-1})^n=a^n(a^{-1})^n=a^ne=a^n$.

So, $a^n=e \iff (a^{-1})^n=e$.

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  • $\begingroup$ Again,this is true only if the elements have finite order,so this doesn't really answer the question. $\endgroup$ Nov 6, 2014 at 6:40
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    $\begingroup$ not really, it clearly implies that $a$ has finite order if and only if $a^{-1}$ has finite order. $\endgroup$
    – Asinomás
    Aug 7, 2016 at 21:19
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    $\begingroup$ $\left(aa^{-1}\right)^n=a^n\left(a^{-1}\right)^n$ if and only if $G$ is abelian, right? $\endgroup$
    – Traveler
    Feb 19, 2019 at 3:46
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    $\begingroup$ @Traveler no, because $a$ and $a^{-1}$ commute $\endgroup$
    – Asinomás
    Feb 19, 2019 at 15:13
  • $\begingroup$ @Jorge Fernández Hidalgo because they commute means? I am not following what you said. Can you please elaborate? $\endgroup$
    – Darshan
    Apr 23, 2020 at 12:46
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Suppose that $a$ has infinite order. We show that $a^{-1}$ cannot have finite order. Suppose to the contrary that $(a^{-1})^m=e$ for some positive integer $m$. We have by repeated application of associativity that $$a^m (a^{-1})^m=e.$$ It follows that $a^m=e$.

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  • $\begingroup$ But this proof doesn't really answer the question either since all it proves is that a and a-1 both have infinite order-it doesn't show they have the SAME order,which is what the claim is trying to prove in the case where the order isn't finite! $\endgroup$ Nov 6, 2014 at 6:43
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    $\begingroup$ OP took care of the case $a$ and $a^{-1}$ have finite order. So it remains to deal with infinite order. The proof above shows that if $a$ has infinite order so does $a^{-1}$. The same argument applied to $a^{-1}$ shows that if $a^{-1}$ has infinite order so does $a$. Two elements of infinite order have the same order. $\endgroup$ Nov 6, 2014 at 6:56
  • $\begingroup$ Ok,sorry,I missed something,I see it now. My bad. I have an alternate proof below. $\endgroup$ Nov 6, 2014 at 6:59
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Let's say a is an element and n is its order ,then $$a^n=e$$ Repeatedly multiplication by $a^{-1}$ n times $$(a^{-1})^{n}•(a)^{n}=e•(a^{-1})^n$$ $$(a^{-1}•a)^{n}=(a^{-1})^n$$ $$e^n=(a^{-1})^n=e$$ Hence "a" 's inverse is also having order of n.

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    $\begingroup$ You can only go from line 2 to 3 if $a$ and $a^{-1}$ commute. (Which they do.) $\endgroup$ Mar 6, 2019 at 5:08
  • $\begingroup$ This answer merely shows that the order of $a^{-1}$ divides $n$. You also need to demonstrate that $k < n \implies (a^{-1})^k \neq e$. $\endgroup$ Mar 6, 2019 at 5:11
  • $\begingroup$ You only proved that the order of $a^{-1}$ was smaller than the order of $a$ $\endgroup$
    – Julien
    Sep 15, 2023 at 12:53
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It seems a more straightforward solution exists?

If $g$ has infinite order then so does $g^{-1}$ since otherwise, for some $m\in\mathbb{Z}^+$, we have $(g^{-1})^m=e=(g^m)^{-1}$, which implies $g^m=e$ since the only element whose inverse is the identity is the identity. This contradicts that $g$ has infinite order, so $g^{-1}$ must have infinite order.

If $g$ has finite order $n$, then by existence of inverses in a group $$g^n=e\iff$$ $$g^n \cdot (g^{-1})^n=e\cdot(g^{-1})^n\iff$$ $$g^n\cdot g^{-n}=(g^{-1})^n\iff$$ $$ e = (g^{-1})^n$$ This implies $|g^{-1}|\leq n$.

If $|g^{-1}|<n$, say $m$, then $(g^{-1})^m=e=(g^m)^{-1}\implies g^m=e$, which contradicts that $|g|=n>m$. So $|g^{-1}|=n$ if $|g|=n$.

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