Query about a proof of infinitude of primes via Lagrange's theorem

Question

Here is the beginning of the proof as it appears on wikipedia:

Assume $p$ is the largest prime. Any prime divisor of $2^p-1$ satisfies $2^p \equiv 1$ (mod q) meaning that the order of 2 in $U(q)$ is $p$.

What I don't understand is how do we know that the order of $2$ is $p$. I mean if, for example, $g \in G$ has order $3$ then $g^6 = e$ and yet the order isn't 6.

In general, if $g^k=e$ in a group, then we know that the order of $g$ divides $k$. Primes don't have a lot of divisors. — lulu, Jun 14 '23 at 10:38
Dupe, proved as part of $q\mid 2^p-1\Rightarrow p\mid q-1$ [order divides prime $p \Rightarrow$ order $= p$ or $1$] — Bill Dubuque, Sep 14 '23 at 19:05

score 3 · Accepted Answer · answered Jun 14 '23 at 10:47

3

In general, for an arbitrary group $G$ and $g\in G$ we have that if $g^n=e$ then the order $|g|$ divides $n$.

In your case $p$ is prime. And so either $|g|=1$ (i.e. $g$ is neutral) or $|g|=p$.

answered Jun 14 '23 at 10:47

freakish

42,851

I'm not sure why this was downvoted (+1) – Shaun Jun 14 '23 at 14:42

Query about a proof of infinitude of primes via Lagrange's theorem

1 Answers1