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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?

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    $\begingroup$ 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 $\endgroup$ Commented Apr 7, 2021 at 2:57
  • $\begingroup$ 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. $\endgroup$ Commented Apr 7, 2021 at 2:58
  • $\begingroup$ Looks very good to me, barring things already mentioned. $\endgroup$ Commented Apr 7, 2021 at 3:13
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    $\begingroup$ 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. $\endgroup$
    – user861776
    Commented Apr 7, 2021 at 3:59
  • $\begingroup$ 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 $n$. $\endgroup$ Commented Apr 7, 2021 at 16:28

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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.

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