Prove that infinite group must have an infinite number of subgroups.

Question

My teacher told me to prove this using the proof by contradiction. So, I am going to assume that there's an infinite group with finitely many subgroups. Then this will lead to finding at least one element with an infinite order, from which I can derive a contradiction.

But I am lost how I can prove this...

I am thinking if the group has its subgroups, then every element in those subgroups must also have infinite order as they are in their original group, aren't they? So, my conclusion is that every element in infinite group's subgroups must have infinite order except identity element.

But how does this contradict the group is infinite?

Answer 1

If $G$ contains an element $x$ of infinite order, then you're done.

If all the elements of $G$ have finite order, then pick one, say $x$. Then $H = \{1_G, x, x^2, ... \}$ is a finite set as well as a subgroup of $G$. Since $G$ is infinite, you can find a $y\in G$ which is not in $H$. Then $H' = \{1_G, y, y^2, ... \}$ is a finite set as well as a subgroup of $G$. It is different from $H$. Again since $G$ is infinite, you can find a $z \in G$ which is not in $H$ or $H'$. Now $H'' = \{1_G,z, z^2, ... \}$ is a finite set as well as a subgroup of $G$, different from $H$ and $H'$. And so on.

Answer 2

Let $G$ be an infinite group. For $x\in G$ let $G(x)=\{x^j:j\in \Bbb Z\}.$

(I). If $G(x)$ is an infinite set for some $x\in G$ then for $n\in \Bbb N$ let $H(n)=\{x^{nj}:j\in \Bbb Z\}.$ Then $H(n)$ is a sub-group of $G$ and $n\ne n'\implies H(n)\ne H(n').$

(II). If $G(x)$ is finite for every $x\in G $ then $S=\{G(x):x\in G\}$ is an infinite set of subgroups of $G$. Because if $T$ is a finite subset of $S$ then $T=\{G(x): x\in T'\}$ for some finite $T'\subset G,$ so $\cup T$ is finite, so there exists $y\in G\setminus \cup T,$ so $G(y)\not \in T.$