If a group $G$ has only finitely many subgroups, then show that $G$ is a finite group. [duplicate]

If a group $G$ has only finitely many subgroups, then show that $G$ is a finite group. I have no idea on how to start this question. Can anyone guide me?

Suppose $G$ were infinite. If $G$ contains an element of infinite order, then _.

Otherwise, every element of $G$ has finite order, and if $x_1, x_2, \ldots, x_n$ are any finite set of elements of $G$, then the subgroups $\langle x_i \rangle$ cover only finitely many elements of $G$. Therefore __.

I think the negative form of that is easier to prove: "If a group $G$ is infinite, then it has infinitely many subgroups".

What I would then do is take a member $a\in G$ such that $\left< a \right>$ is infinite, then prove there are an infinite number of subgroups just there, and say that if no such $a$ exists, there must be infinitely many subgroups of the type $\left< a \right>$.

What is a "secular equation"?

If series $\sum a_n$ is convergent with positive terms does $\sum \sin a_n$ also converge?

If $f: M\to M$ an isometry, is $f$ bijective?

Without Taylor series $\lim\limits_{n\to\infty}\frac{n}{\ln \ln n}\cdot \left(\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1\right) $

What are some difficult integrals done by substitution and elementary functions?

How to prove that $\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin{t}}{t}\right)^2\,\mathrm dt\le 1-\dfrac{1}{p}$ for $p >1, x\ge0$

Let $f :\mathbb{R}→ \mathbb{R}$ be a function such that $f^2$ and $f^3$ are differentiable. Is $f$ differentiable?

Can we smoothly embed $\mathbb{S}^2 \times \mathbb{S}^1$ or $\mathbb{RP}^2 \times \mathbb{R}$ in $\mathbb{R}^4$?

Is there a better alternative to the phrase, 'it holds that'?

Do isomorphic structures always satisfy the same second-order sentences?

What determines if a function has a least positive period?

injection $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$