# Prove that in any group, an element has the same order of its inverse

• Dec 3rd 2010, 11:05 AM
SunRiseAir
Prove that in any group, an element has the same order of its inverse
Am I on the right track?

This will be a proof by contradiction.

i. Assume $a$ has order n. Therefore,
$a^n = e$
ii. Likewise, assume that $a^{-1}$ has order m, with m different from n. Say, $m < n$. Therefore,
$a^{-m} = e$
iii. Since $a^n = e$ and $a^{-m} = e$
$a^n = a^{-m}$

Multiply both sides by $a^{-m}$ on the right. Then,

$a^n a^{-m} = a^{-m} a^{-m}$

By laws of exponents,

$a^{n-m}= (a^{-m})(a^{-m})$

By the definition of order, on the right we have that
$(a^{-m}) (a^{-m}) = e$
So then we have,

$a^{n-m}= e$
iv. Since $m < n$, as defined at step ii., it follows that
$n - m < n$
v. But the result of iv. contradicts the claim that a has order n, since by definition this means that n is the least positive integer such that $a^n = e$ holds.

vi. Therefore an element has the same order of its inverse.

Thanks for anyone who helps! (Itwasntme)
• Dec 3rd 2010, 11:42 AM
emakarov
Formally, you also need to consider the case when m > n, but I agree with your proof so far.

It may be easier to just note that $(a^n)^{-1}=a^{-n}$ for all integer n. Therefore, $a^n=e$ iff $a^{-n}=e$. So, $\{n\in\mathbb{Z}\mid a^n=e\}=\{n\in\mathbb{Z}\mid (a^{-1})^n=e\}$. From here the claim follows.