Proof that if $G$ is a group with no proper subgroups and order $>1$, then $G$ is cyclic with prime order

Question

As the title states, I want to prove that if $G$ is a group with no proper subgroups and order at least 2, then $G$ is cyclic with prime order. This has been proven before in here, but I want to know if my own proof and explanations are correct.

Proof: Since $|G|>1$, we know that $\exists g∈G :|g|>1$. Consider the subgroup $⟨g⟩⊆G$ which is cyclic by definition. Since $G$ has no proper subgroups, we must have $⟨g⟩=G$. It is now to be proven that $|g|=p$ with $p=$ prime. I will show it in two steps, both by contradiction (edit: as pointed out, they are really proofs by contraposition): 1) $$ is finite. 2) $||=$ with $=$ prime.

1. Assume that $|g|=\infty$. Now, $g^2\in\langle g\rangle$ so $\langle g^2\rangle$ is a subgroup of $\langle g\rangle$. Because $g^{n} \notin\left\langle g^{2}\right\rangle$ whenever $n$ is odd, $\langle g^2\rangle$ is a proper subgroup of $\langle g\rangle$. This contradicts the assumption that $G$ has no proper subgroups, so $g$ must be finite.
2. Define the order of $g$ by $n$ and assume that $n$ is not prime, i.e., $\exists d\in \mathbb{Z} :d|n$ with $1<d<n$. Because $g^d\in\langle g\rangle$, we know that $\langle g^d\rangle $ is a subgroup of $\langle g\rangle$. Additionally, a theorem tells us that $|g^d|=n/d<n$ so $\langle g^d\rangle$ is a proper subgroup of $\langle g\rangle$. This is again a contradiction, so $n$ must be prime.

Is my proof correct and sufficient?

Answer 1

Your argument is correct. I will point out that your argument in 1 is what I sometimes call a "proof by fake contradiction" or "fake proof by contradiction" (see discussion here). They are really proofs by contrapositive: you are showing that if the order $g$ is not a prime, then $G$ will not satisfy the hypotheses.

You can turn your argument into a direct proof that $g$ must have finite order:

Let $g\neq e$. Then $\langle g\rangle = G$. Now, $\langle g^2\rangle$ is either trivial (in which case $g$ has order $2$ and you are done); or else $\langle g^2\rangle = G = \langle g\rangle$. In this case, $g\in\langle g^2\rangle$, so there exists $k$ such that $g=(g^2)^k = g^{2k}$, hence $1=g^{2k-1}$ shows $g$ has finite order.

The second part can likewise be done directly: let $n=|g|$, which we now know is finite. If $1\lt d\lt n$, then $\langle g^d\rangle = \langle g\rangle$, so there exists $k$ such that $g=g^{dk}$. Then $n\mid dk-1$, so $\gcd(n,d)=1$. Thus, $n$ is relative prime to all integers $d$ with $1\lt d\lt n$, so we conclude that $n$ is prime, as required.