If G has no proper subgroups, prove that G is cyclic of order p, where p is a prime number.
Assume that G has no proper subgroups, i.e., if H is a subgroup of G, then H={e} or H=G.
If G={e}, then {e} is clearly cyclic. (e)={e}.
If G=/= {e}, then ther eis an a in G with a =/= e. Then (a) is a subgroup of G, (a) =/= {e}. Then our assumption that G has no proper subgroups implies that (a)=G, so G is cyclic. Actually, the cyclic subgroup of any element of G is equal to G (since there can be no proper subgroups).
Then (a) is a subgroup of G, where (a)=/={e}. Then our assumption that G has no proper subgroups, implies that (a)=G, so G is cyclic.
For contradiction, assume that such a group is of nonprime order. So G={e,a,a^2,...,a^(n-1)} where n=qb with q,b in Z, and where q and b are not equal to 1 or n (since n is not prime, this must be possible). Since b < qb-1, a^b must be an element of G. According to the assumption, (a^b)=G.
(a^b) = {e, a, a^2, ... , a^(n-1)}. So there must be an element k in G such that kb= qb-1 (mod n) (this is of course a congruence sign, not an equal sign).
So n divides qb-1-kb. So there is an integer c such that
qb-q-kb= nc
qb-1-kb=(qb)c
b(q-k)-1=c(qb)
We know that b(q-k)-1 < b(q-k) < qb < c(qb). So b(q-k)-1 < c(qb).
So the equality sign is not true, and this is a contradiciton.
Is my proof correct? If not, can you please tell me why?
Thanks in advance