I'm studying Cauchy theorem in Group Theory and I found an interesting proof in the book Visual Group Theory. At the beginning of this proof, there's a claim which puzzles me a bit (here G is a finite group, $g\in G$ and p is a prime):
"Because p is prime, if I find some $g\ne e$ satisfying $g^p=e$, then $g$ must have order p."
Actually what make me think is the definition of the order of an element is precisely "a number n which makes $g^n=e$, and this no matter if n is prime or not. So that sentence it seems to me a bit over-constrained, since it would be true no matter if p is prime or not.
Where am I wrong in my reasoning, please?
Thank as always for your support.
No, in general it does not follow from $g^n=e$ that $g$ has order $n$, because already $g^{n/d}=e$ could hold for a divisor of $n$. However, if $n=p$ is prime, then of course it does follow. As an example, take $G=GL_2(\Bbb Z)$ and $$ g=\begin{pmatrix} 1 & 0 \cr 0 & -1\end{pmatrix}. $$ Of course, $g^4=e$, but the order of $g$ is $2$.