Show that a group that has only a finite number of subgroups must be a finite group.

I started by assuming that group is infinite.

But, don't understand how I should go from this assumption

  • 2
    $\begingroup$ Start by proving that every element has finite order. $\endgroup$ Feb 4, 2015 at 5:22

1 Answer 1


Note that each $a\in G$ generates a subgroup $\langle a\rangle$. Noting that

$$G=\bigcup_{a\in G}\langle a\rangle$$

and any elements of infinite order produce a subgroup isomorphic to $\Bbb Z$, so that this subgroup has infinitely many subgroups--i.e. if $\langle a\rangle$ is infinite, then $\langle a^n\rangle, n\in\Bbb N$ is an infinite family of subgroups (a contradiction). But then $G$ is a finite union of finite sets, making it a finite set itself.

  • $\begingroup$ Are you saying that every element in G is generating subgroup of G? $\endgroup$
    – nany
    Feb 4, 2015 at 5:48
  • $\begingroup$ @nany yes, all elements generate subgroups $\endgroup$ Feb 4, 2015 at 5:49
  • $\begingroup$ Sorry, but Could you explain why all elements generate subgroups..? $\endgroup$
    – nany
    Feb 4, 2015 at 5:53
  • $\begingroup$ @nany by definition, just look at the set $\{a^n : n\in\Bbb Z\}$. This is a subgroup as $a^n*_Ga^m=a^{n+m}$ and $(a^n)^{-1}=a^{-n}$ and $a^0=e_G$. $\endgroup$ Feb 4, 2015 at 5:54
  • $\begingroup$ @nany: For any group $G$ and any subset $S \subset G$, the subgroup of $G$ generated by $S$, denoted $\langle S \rangle$, can be defined. One definition of $\langle S \rangle$ is the set of all products of finite sequences of elements of $S$ and their inverses. Another, equivalent definition is the intersection of all subgroups of $G$ that contain $\langle S \rangle$. $\endgroup$
    – Lee Mosher
    Feb 4, 2015 at 17:53

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