Proving Finite Groups with No Non-Trivial Subgroups and Prime Order o(G)

[QuArK21343]

Prove that if a group G has no non-trivial subgroups, G is finite and o(G) is a prime number, where o(G) is the order of the group G.

If G is infinite, you can show that there are non trivial subgroups. What remains to prove is that if o(G) is not prime, than there is at least one subgroup H, with o(H) equal to one of the prime divisor of o(G). Any idea?

 

[micromass]

Suppose that G has no nontrivial subgroups. Take an arbitrary element x with x\neq e. Then what can you say about <x> (the group generated by x)??

 

[QuArK21343]

Since G has no nontrivial subgroup and <x> is a subgroup of G, but <x> is not e, G=<x>.

 

[micromass]

Since G has no nontrivial subgroup and <x> is a subgroup of G, but <x> is not e, G=<x>.

So you must only prove now that all cyclic subgroups whose order is not prime have a nontrivial subgroup...

 

[QuArK21343]

I prove that if G=<g> is a finite cyclic group and o(G)=m, then for every d such that d|m there is a subgroup H such that o(H)=d. It is simply this: consider the subgroup <g^(m/d)>. Since o(G)=o(g)=m, we have that o(H)=o(g^(m/d))=m/(m/d)=d. So, H is a non trivial subgroup. So, if the only possible subgroups are the trivial ones, m must be prime.

 

[micromass]

Seems ok!