Congruence through Groups

Question

I was reading proof and have gotten stuck in a place. I needed some help understanding a statement they have made:

Let $G$ be a group of order $pqr$. Let $P,Q,R$ be Sylow subgroups of $G$ of order $p,q,r$ respectively. Let $\nu_p,\nu_q,\nu_r$ denote the number of Sylow subgroups of order $p,q,r$ respectively. Then, they have proved that at least one of $\nu_q,\nu_r$ must be $1$. I have understood this.

Let that subgroup be $X=\langle x \rangle$ and it will be normal to $G$ by Sylow's Theorem on conjugates. Let $P=\langle a \rangle$ ($P$ must be of order $p$ by Lagrange's theorem and hence cyclic. Similar reasoning for $X$.) Suppose $X$ is of order $l$.

Let $axa^{-1}=x^i.$ Then, $x=a^pxa^{-p}=x^{i^{p}}$. Hence $i^p \equiv 1 \textrm{ mod }l$. Since, $i^{l-1}\equiv 1 \textrm{ mod }l...$.

I did not understand how they arrived at the last statement, i.e., "Since, $i^{l-1}\equiv 1 \textrm{ mod }l$. " Please help. !

Answer 1

This is a consequence of Fermat' little theorem: if the integer $a$ is not divisible by the prime $p$, $$a^{p-1}\equiv 1 \mod p.$$ Then just see that $l$ must be a prime and, by Lagrange's Theorem, $i$ does not divide $l$.

Answer 2

Just rewrite $x=x^{i^{p}}$ as $x^{i^{p}-1}=e$ (I denote $e$ the unit element in the group).

As $x$ has order $l$, this means that $\;i^{p}-1\equiv 0\mod l$.

As to $i^{l-1}$, observe that $l$ denotes the order of one of the Sylow subgroups $Q,R$ which is normal? These subgroups have prime orders $q$ and $r$. Thus, whatever $X$ is, its order is a prime ($q$ or $r$), denoted $l$. As $i<l$, lil' Fermat asserts that $i^{l-1}\equiv 1\mod l$.