If $G$ has no non trivial subgroups, then show that $G$ must be of prime order. If $G$ has no non trivial subgroups, then Show that $G$ must be of prime order. This question is from Herstein Page 46 Question 3.
Attempt:
Let $G$ has prime order(say $p$). By Lagrange theorem, order of the subgroup can be $1$ or $p$. So it has two trivial subgroups.
How do i prove the other way around?
Thanks.

Consider the subgroups generated by one element - that is, those of them form $\{e,x,x^2,x^3,\ldots\}$. Now, if $G$ is to have no non-trivial subgroups, then if we choose some $x$ other than the identity, this subgroup cannot be the trivial group - thus, for it to be a trivial subgroup, it's got to be the whole group $G$ (which, as an aside, means $G$ is a cyclic group). Now, to finish, consider: if $|G|=ab$, what can we say about the subgroup generated by $x^a$?
