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

• Dec 3rd 2010, 10:05 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 $\displaystyle a$ has order n. Therefore,
$\displaystyle a^n = e$
ii. Likewise, assume that $\displaystyle a^{-1}$ has order m, with m different from n. Say, $\displaystyle m < n$. Therefore,
$\displaystyle a^{-m} = e$
iii. Since $\displaystyle a^n = e$ and $\displaystyle a^{-m} = e$
$\displaystyle a^n = a^{-m}$

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

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

By laws of exponents,

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

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

$\displaystyle a^{n-m}= e$
iv. Since $\displaystyle m < n$, as defined at step ii., it follows that
$\displaystyle n - m < n$
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 $\displaystyle a^n = e$ holds.

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

Thanks for anyone who helps! (Itwasntme)
• Dec 3rd 2010, 10:42 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 $\displaystyle (a^n)^{-1}=a^{-n}$ for all integer n. Therefore, $\displaystyle a^n=e$ iff $\displaystyle a^{-n}=e$. So, $\displaystyle \{n\in\mathbb{Z}\mid a^n=e\}=\{n\in\mathbb{Z}\mid (a^{-1})^n=e\}$. From here the claim follows.