$|G|=p(p+1)$ for $p$ prime, then $G$ has a normal subgroup of order $p$ or $p+1$

I am trying to solve the above question, as an application of Sylow's theorem. Let $P$ be the p-Sylow subgroup. Then $n_p | (p+1)$ and $n_p \equiv 1 \pmod{p}$. If $n_p =1$, $P$ is normal and we are done, else $n_p = p+1$. Now,

\begin{equation} 1+n_p(p-1) = 1 + (p+1)(p-1) = p^2, \end{equation} is the total number of elements in the p-Sylow subgroups. So if $n_p = p+1$, that means the number of elements not in any p-Sylow subgroup is $|G|-p^2=p$. If these $p$ elements and the identity form a subgroup then its a subgroup of the smallest prime index, so we are done.

But how do I show that all the elements not in the p-Sylow subgroups form a subgroup, i.e. the subgroup generated by these elements has trivial intersection with the $p-$Sylow subgroups?

Solution 1:

I will expand my comment into an answer.

Let $S$ be the set of elements that do not lie in any Sylow $p$-subgroup of $G$. You have shown by a counting argument that $|S|=p$. Let $q$ be any prime that divides $p+1$.Then $S$ must contain some element $g$ of order $q$.

Since $n_p=p+1$, we have $N_G(P) = P$, so $g \not\in N_G(P)$. Let $x$ be a generator of $P$. Then the powers $x,x^2,\ldots, x^{p-1}$ of $x$ all generate $P$, so none of them can centralize $g$. Hence the $p$ elements $\{ g, g^x,g^{x^2}, \ldots, g^{x^{p-1}} \}$ (where $g^h$ means $hgh^{-1}$) are all distinct. Since they all have order $q$, they all lie in $S$, and so $S = \{ g, g^x,g^{x^2}, \ldots, g^{x^{p-1}} \}$.

So every element of $S$ has order $q$, and hence $q$ must be the only prime dividing $p+1$, so $p+1$ is a power of $q$, and $S \cup \{ 1 \}$ must be the unique Sylow $q$-subgroup of $G$.

Solution 2:

You can apply Burnside's p-complement theorem, since $P=C_G(P)=N_G(P)$, $P$ has a normal complement $N$. So $G=PN$ and $P \cap N =1$ and hence $|N|=p+1$.