If a group has no subgroups other than the identity and itself, then it is finite and is of prime order

Question

I want to prove that if a group has no subgroups other than the identity and itself, then the order of the group is a prime number.

A hint would be appreciated. Is there any theorem on the relation of the order of the subgroup and that of the group?

Note that the finiteness is not given, unlike in the linked question.

The order of a subgroup must divide the order of the group by Lagrange's theorem. — Sri-Amirthan Theivendran, Feb 07 '16 at 23:20
@FoobazJohn is Lagrange's theorem for finite groups? I had not read that theorem, but seems like it is about finite groups. In my case finiteness is not given — Frank Booth, Feb 07 '16 at 23:22
If G has at least three elements it has an a,b such that a $\ne$ b and neither b nor a are equal to the identity. Then and are subgroups. So = = G and a = b^m and b = a^n and a^mn = a so = G is finite of order equal or less than nm-1. So G isn't infinite. — fleablood, Feb 08 '16 at 01:17
The case of an infinite group is also covered in the linked thread. I am fairly sure that the linked thread was not the first time this question was handled, but back in the day such things were only considered worthy of a comment :-/ — Jyrki Lahtonen, Feb 08 '16 at 08:11

Eric Thoma · Answer 1 · 2016-02-07T23:36:19.853

First we need to prove that $G$ is finite. Note that every nontrivial element generates the group. Take a generator $x$ of $G$. Then if $G$ has infinite order, we have that $x^n \ne x$ for any $n > 1$. But $x^2$ is a generator as well, and so we have $x^{2n} = x$ for some $n \geq 1$, contradiction. We conclude that $G$ is finite.

Suppose $G$ has composite order, and so its order is $r \cdot s$ for $r,s \ne 1$. Take a generator $x$ of $G$, and note that $x^r$ has order $s$. But then $\langle x^r \rangle$ is a nontrivial subgroup, contradiction.

If a group has no subgroups other than the identity and itself, then it is finite and is of prime order

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