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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?

Jendrik Stelzner
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ChocolateDroppa
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    Note that this is exercise 4 in chapter 3 in Gallian's Contemporary Abstract Algebra. – a student Dec 22 '15 at 11:05
  • 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). – j.p. Nov 09 '21 at 06:59
  • See also https://math.stackexchange.com/questions/2794098/let-g-be-a-group-and-a-belongs-to-g-prove-that-langle-a-1-rangle – lhf Nov 10 '21 at 10:23

4 Answers4

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

Christopher Creutzig
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Asinomás
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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 $a^m=e$.

André Nicolas
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  • 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! – Mathemagician1234 Nov 06 '14 at 06:43
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    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. – André Nicolas Nov 06 '14 at 06:56
  • Ok,sorry,I missed something,I see it now. My bad. I have an alternate proof below. – Mathemagician1234 Nov 06 '14 at 06:59
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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.

amWhy
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user496040
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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 $|g|=n$.

coreyman317
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