# $|G|=p(p+1)$ 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,

$$1+n_p(p-1) = 1 + (p+1)(p-1) = p^2,$$ 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?

• An element of order $p$ cannot centralize any of these $p$ elements because $P$ is self-normalizing, so the $p$ elements are all conjugate, so they must all have the same prime order $q$, and now we see that they and the identiity form a Sylow $q$-subgroup. – Derek Holt Oct 4 '16 at 15:55
• @DerekHolt what is $q$ – basket Oct 4 '16 at 15:59
• The order of each of the $p$ elements that do not lie in a Sylow $p$-subgroup. – Derek Holt Oct 4 '16 at 16:03

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 $G$.

• nice answer, however I don't get what do you mean by the only prime dividing $p+1$? This is an even number, so there is for sure one prime dividing it, namely 2. Also you mean that $S$ is a subgroup if we add the trivial element of $G$ right? Am I wrong somewhere? – user321268 Oct 4 '16 at 20:19
• Yes I meant $S$ together with the identity. But what is unclear about "the only prime dividing $p+1$"? It means exactly what it says. Yes in fact $q=2$. – Derek Holt Oct 4 '16 at 20:27
• Yes that's right. In fact for all Mersenne primes $p$ there is a unique isomorphism type of groups with this property. – Derek Holt Oct 4 '16 at 20:49
• @RFZ Have you read the definition of $S$? We know that $q \ne p$, so how could an element of order $q$ be contained in a Sylow $p$-subgroup? Your second question follows immediately from Sylow's theorem. – Derek Holt Mar 14 '18 at 8:16
• The existence of an element of order $q$ follows from Cauchy's theorem, which itself follows from Sylow's theorem. It seems clear that people attempting this problem are expected to be familiar with Sylow's Theorem. – Derek Holt Mar 14 '18 at 9:22

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 $|N|=p+1$.

• that's for sure a correct answer, though, I was wondering if there is any other alternative proof Nicky. I'm not sure that someone undergraduate for instance, is familiar with that theorem cause seems to be a more elementary exercise. – user321268 Oct 4 '16 at 18:25
• Ok fair enough, I cannot read off the OP post the level. Then better follow Derek's line of reasoning. – Nicky Hekster Oct 4 '16 at 19:02