Is G a Cyclic Group of Prime Order?

[semidevil]

let G be a non-trivial group whose only supgroups are {e} and G. Show that G is cyclic of prime order.

ok...so basically, it is asking I need show that it is cyclic that has an order that is only divisible by 1 and itself right?

I don't know how to approace this, but I know the Fundamental Theorem of Cyclic groups, which says "Every subgroup of cyclic group is cyclic. So if the only subgroups are the identity and itself, by the problem, doesn't that mean that it is prime order?

 

[Hurkyl]

(a) Suppose it's not cyclic. What can you prove?
(b) Suppose it's not of prime order. What can you prove?

 

[wais]

Yes, you are correct. Since G is a non-trivial group, it must have at least one non-identity element. Let's call this element g. Since G is a cyclic group, there exists an integer n such that g^n = e (the identity element). This means that the order of g is n.

Now, let's consider the subgroup generated by g, denoted <g>. By the Fundamental Theorem of Cyclic groups, <g> is also a cyclic group. But since the only subgroups of G are {e} and G itself, <g> must be equal to G. This means that every element in G can be written as a power of g, and thus G is cyclic.

Since the order of g is n, and every element in G can be written as a power of g, the order of G must also be n. But since n is the order of a non-identity element in G, it cannot be divisible by any number other than 1 and itself. Therefore, G is cyclic of prime order.