A group element and its inverse have the same order

Let G be a group and let x \in G. Prove that x and x^{-1} have the same order.


Recall that the order of a group element is either a positive integer or infinity.

Suppose |x| is infinite and that |x^{-1}| = n for some n. Then x^n = x^{-1 \cdot n \cdot -1} = ((x^{-1})^n)^{-1} = 1^{-1} = 1, a contradiction. So if |x| is infinite, |x^{-1}| must also be infinite. Likewise, if |x^{-1}| is infinte, then |(x^{-1})^{-1}| = |x| is also infinite.

Suppose now that |x| = n and |x^{-1}| = m are both finite. Then we have (x^{-1})^n = (x^n)^{-1} = 1^{-1} = 1, so that m \leq n. Likewise, n \leq m. Hence m = n and x and x^{-1} have the same order.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: