Algebra Solutions: 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 $|x| \leq |G|$ .

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 $|x| = n \leq |G|$ .

Algebra Solutions

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

No comments:

Post a Comment