Groups order 957 are cyclics

Question

Let G be an order 957 group. I have to prove that it is cyclic.

When I calculate Sylow p-subgroups for p=3, I find that 319 is a valid number or Sylow 3-subgroups. p=11 and p=29 gives me a normal Sylow p-subgroup for each of them.

I know that, if all Sylow groups were normal, we could probe the isomorfism to cyclic 957 group. By Sylow, how could this be solved?

But there isn't normal subgroups order 3, so I don't know how to prove that it's cyclic. — Pablo Luis, Jan 12 '18 at 19:58
Yes, we have $\phi(957)=560$, which is coprime to $957$. So we are done. This uses the Sylow theorems, of course, see here. — Dietrich Burde, Jan 12 '18 at 20:02

score 0 · Answer 1 · answered Jan 12 '18 at 20:24

0

A quick look here tells us that $957$ is coprime to $\phi(957)$, where $\phi$ is the totient function. So we can use the strategy discussed here.

answered Jan 12 '18 at 20:24

BallBoy

14,472

Groups order 957 are cyclics

1 Answers1