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

**SunRiseAir** · December 3rd 2010, 10:05 AM

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!

**emakarov** · December 3rd 2010, 10:42 AM

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.

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

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

Inverse of an element in a group

order of an element of a group

If a is an element of order m in a group G and a^(k) = e, prove that m divides k.

Prove that the identity element e is the only idempotent element in a group G

group element order

Search Tags