I am currently attempting to understand the proof of
Proposition 2.13. In a finite group, the number of elements of prime order $p$ is divisible by $p − 1.$
From Carrell's "Groups matrices and vector spaces".
The proof is as follows:
By Lagrange, two distinct subgroups of prime order $p$ meet exactly at $1$, for their intersection is a subgroup of both. But in a group of order $p$, every element except the identity has order exactly $p$, since $p$ is prime. Thus the number of elements of order $p$ is $m(p − 1)$ where $m$ is the number of subgroups of order $p.$
I am having difficulty understanding why every element in the subgroups has order $p.$
I am aware that groups with prime order are cyclic. So does this imply that every element of a prime order cyclic group, except the identity, have order $p$? And if so why?
Best Answer
Let $o(G) = p$. if $a\in G - \{1\} $ then the group generated by $a$ is a subgroup of $G$, therefore its order divides $p$ and therefore it is $p$ (it cannot be 1 because then $a = 1$). Because the order of an element is the order of the group it generates, we have that every element except for the identity is of order $p$.