If you find any mistakes, please make a comment! Thank you.

The order of a group element is smaller than the cardinality of the group


Let $G$ be a group and $x \in G$ an element of order $n < \infty$. Prove that the elements $1, x, x^2, \ldots, x^{n-1}$ are all distinct. Deduce that $|x| \leq |G|$.


Solution: Suppose to the contrary that $x^a = x^b$ for some $0 \leq a < b \leq n-1$. Then we have $x^{b-a} = 1$, with $1 \leq b-a < n$. However, recall that $n$ is by definition the least integer $k$ such that $x^k = 1$, so we have a contradiction. Thus all the $x^i$, $0 \leq i \leq n-1$, are distinct. In particular, we have $$\{ x^i \ |\ 0 \leq i \leq n-1 \} \subseteq G,$$ so that $|x| = n \leq |G|$.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu