# Proof showing that a group must be finite of prime order

I'm trying to prove the following:

G is a group with order $\ge 2$ with no proper, non-trivial subgroups. G must be finite of prime order.

My attempt:

Consider $g \neq e \in G$ (we can do this since order of $G$ is at least 2). Since $G$ has no proper, non-trivial subgroups, $<g>$ can't be a proper subgroup of $G$. Since it clearly can't be $e$, we must have $<g> = G$.

I'm not sure why it has to be finite though...

Help?

Thanks guys, Mariogs

• Show that if $g$ generates $G$, and $G$ isn't finite, then $2g$ generates a proper subgroup of $G$. – Gerry Myerson Mar 8 '14 at 22:15

## 3 Answers

So you have $\langle g\rangle=G$. If $g$ has infinite order, then $\langle g^2\rangle$ is proper in $G$ for instance, since $g\notin\langle g^2\rangle$. If it where, you'd have $g=g^{2k}$ for some $k$, or $g^{2k-1}=e$, contradiction.

So $G$ is a finite cyclic group. Recall that a finite cyclic group has a unique subgroup of every order dividing $|G|$. This forces $|G|$ to be prime, otherwise you'd have a nontrivial, proper subgroup.

I suppose this question is from Herstein. If You have solved the previous one you know that in a group G if intersection of all it's subgroup different from is a subgroup different from then all of it's elements have finite order. Here indeed this is the case.

• Actually not from Herstein! Thanks for the response though, helpful! – bclayman Mar 9 '14 at 14:58

|G|>=2 then G has a nonidentity element say a . Consider the sub group generated by a is . as G has no nontrivial sub group therefore G= .now G is cyclic . if |G| is infinite then G is isomorphic to Z .but Z has non trivial sub groups .therefore order of G is finite say n .now G to be cyclic ,G has a subgroup for each divisor of n . but G has no nontrivial sub group .therefore divisor has to be 1 or n thus n is prime.

• This proof isn't that different from the one Ben West gave over a year ago. – Michael Albanese Nov 15 '15 at 13:45