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I want to show that there are no simple groups of order $p^{k}(p+1)$ where $k>0$ and $p$ is a prime number.

So suppose there is such a group. Then if we let $n_{p}$ denote the number of $p$-Sylow subgroups of $G$ we have that $n_{p}=p+1$. Now by letting $G$ act on $Sylow_{P}(G)$ by conjugation we obtain a group homomorphism $G \rightarrow S_{p+1}$. Since $G$ is simple then either $ker(f)$ is trivial or all $G$. Now here's my question: assume $ker(f)=G$ this would imply then that $G$ has a unique $p$-Sylow subgroup no? but then such subgroup is normal which contradicts the fact that $G$ is simple. So the map in fact is injective but then $|G|$ divides $(p+1)!$ which cannot be.

Basically my question is if my argument is correct, namely thta if $ker(f)=g$ implies the existence of a unique $p$-Sylow subgroup which implies such subgroup is normal in $G$ which cannot be. In case this is wrong, how do you argue that $ker(f)$ cannot be all $G$?


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up vote 4 down vote accepted

You are (mostly) correct. If ker(f) = G, then the image of G in the symmetric group is just the identity, so G does not move any of its Sylow p-subgroups around. However, G acts transitively on its Sylow p-subgroups (they are all conjugate) and the identity is not transitive unless p+1=1, which is silly.

If ker(f) = 1, then G embeds in the symmetric group on p+1 points, so its order divides (p+1)!. This is not a contradiction when k=1.

Indeed, taking p=2, the non-abelian group of order six has order p(p+1) and has exactly p+1 Sylow p-subgroups, and the homomorphism f is injective.

Of course a group of order p(p+1) is also not simple, but probably for a different reason.

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For a group of order $p(p+1)$ with $n_p=p+1$, there would be $p^2-1$ elements of order $p$, and the remaining $p$ non-identity elements would all be conjugate under the action of a Sylow $p$-subgroup, so they would all have the same order, so $p+1$ would be prime, which is a rare occurrence. – Derek Holt Mar 3 '11 at 20:39
Sorry, I meant of course that $p+1$ would be a prime power $q^r$, and then the $p+1$ elements not of order $p$ would form a unique and hence normal Sylow $q$-subgroup. Either $p=2$, $q=3$, or $q=2$ and $p$ is a Mersenne prime. – Derek Holt Mar 3 '11 at 20:50
@Derek: I like your proof; it just uses Sylow and Lagrange, though I guess the Mersenne prime case is annoying (do you need to use Burnside's pq?). My proof was that |G|=p(p+1) with np=p+1 implies G is p-nilpotent (since its Sylow p-subgroup is self-normalizing and abelian). – Jack Schmidt Mar 3 '11 at 22:16
@Jack: The non-simplicity proof for $p(p+1)$ just uses Sylow, Lagrange and counting. Note also that Im$(f)$ is a 2-transitive Frobenius group. Although the general proof that a Frobenius group has a regular normal subgroup uses character theory, it is easily proved in the 2-transitive case, using the same argument as above. – Derek Holt Mar 4 '11 at 9:15
@Derek: (Using just counting, Sylow, Lagrange, not Frobenius groups or p-nilpotence criteria) When q=2, how do you know the Sylow 2-subgroup is normal? It seems plausible (but impossible) that there are p Sylow 2-subgroups that overlap a lot. – Jack Schmidt Mar 4 '11 at 15:33

"Now here's my question: assume ker(f)=G this would imply then that G has a unique p-Sylow subgroup no? "

yes, the group action (conjugation on sylow subgroups) is transitive.

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