Finding all groups with no proper subgroups

I am trying to solve the following problem in Artin's algebra.

Describe all groups $$G$$ that contain no proper subgroups.

Here is my attempt.

If $$G = \{e\}$$, then it is clear $$G$$ has no proper subgroups, so suppose that $$G \neq \{e\}$$. Let $$x \in G$$ be a non-identity element, and consider $$\langle x \rangle$$, the cyclic group generated by $$x$$. Certainly $$\langle x \rangle \leq G$$, but if $$G$$ has no proper subgroup and $$\langle x \rangle \neq \{e\}$$ since $$x \neq e$$, we have $$G = \langle x \rangle$$. So $$G$$ must be cyclic. We prove that $$G$$ must be finite. Suppose $$G$$ were infinite. Then $$\langle x \rangle \cong \mathbb{Z}$$ by the map $$\varphi: \mathbb{Z} \to \langle x \rangle$$ sending $$n \longmapsto x^n$$. But $$2\mathbb{Z}$$ is a non-proper subgroup of $$\mathbb{Z}$$, and the image of a subgroup under a homomorphism is a subgroup of the codomain. In particular, $$2\mathbb{Z}$$ is the cyclic subgroup of $$\mathbb{Z}$$ generated by $$2$$, so $$\varphi(2\mathbb{Z})$$ is the cyclic subgroup of $$G$$ generated by $$x^2$$. But $$x \not \in \langle x^2 \rangle$$, so $$\langle x^2 \rangle \neq G$$. Furthermore, it is not the identity since $$x$$ has infinite order. So we have found a proper subgroup of $$G$$, a contradiction, so $$G$$ has finite order. Finally, we prove that $$G$$ must have prime order. Suppose the order of $$G$$ is not prime, i.e., $$|G| = ab$$ for $$a,b > 1$$. Let $$G = \langle x \rangle$$, so $$|x| = |\langle x \rangle| = ab$$. Then $$\left(x^a\right)^b = e$$, so $$\langle x^a \rangle$$ is a cyclic subgroup of $$G$$ of order $$b$$; since $$a,b > 1$$, this is a proper subgroup, so we have a contradiction. Hence, $$|G| = p$$, where $$p$$ is prime.

Finally, we must show that all prime cyclic groups contain no proper subgroups. If $$G$$ is a cyclic group of prime order $$p$$, then any subgroup $$H \leq G$$ must have order dividing the order of $$G$$ by Lagrange's theorem. But the only divisors of $$p$$ in $$\mathbb{N}$$ are $$1$$ and $$p$$, i.e., $$\{e\}$$ and $$G$$. So there are no proper subgroups of $$G$$, as desired.

How does this look?

• A typo, $\langle x \rangle \cong \mathbb{Z}$ rather than $G \cong \mathbb{Z}$, ... well at least if $\langle x \rangle$ is infinite then it's isomorphic to the integers, but you could have finite $\langle x \rangle$ in infinite $G$. Small semantic issue though Commented Apr 7, 2021 at 2:57
• You have to say more to justify that $\langle x^a\rangle$ has order $b$; the only thing you proved is that it has order dividing $b$, not that it has order exactly $b$. There is also no need to argue by contradiction that $G$ is finite: since $G$ has no proper subgroups, either $\langle x^2\rangle = G$ or $\langle x^2\rangle$ is trivial. In the latter case, $G$ has order $2$. In the former case, $x\in\langle x^2\rangle$, so there exists $k$ such that $x = (x^2)^k = x^{2k}$, proving $x$ has finite order. Commented Apr 7, 2021 at 2:58
• Looks very good to me, barring things already mentioned. Commented Apr 7, 2021 at 3:13
• This is extremely helpful, thank you. Do you have a hint for proving that $x^a$ has order exactly $b$? Should I proceed by contradiction, or prove directly that $b \mid |x^a|$? I'll update the first post as I work more on this. Commented Apr 7, 2021 at 3:59
• You really only need to show that $x^a\neq e$, and has order strictly smaller than $n$, in order to show that $\langle x^a\rangle$ is a proper nontrivial subgroup. In fact, you are already arguing by contradiction, so adding yet another argument by contradiction inside your argument by contradiction is counter-indicated. You could prove $|G|$ is prime directly as follows: let $1\lt d$ be a divisor of $n=|G|$. Then the order of $x^d$ divides $n/d\lt n$, so $\langle x^d\rangle$ is a proper subgroup of $G$, hence trivial. So $x^d=e$, and $n|d$. Thus, $n=d$, so the only divisors are $1$ and $n$. Commented Apr 7, 2021 at 16:28

Your proof will go through, with the adjustment in the first part that $$|x^a|\mid b$$. That's sufficient. Since then we get a proper subgroup. You can prove $$|x^a|=b$$, but it isn't necessary.