Does every finite group of size at least 2 has finite cyclic subgroup of size at least 2?

Lagrange's theorem doesn't help, because it doesn't say anything about existence of subgroup of some order.

  • 4
    $\begingroup$ What about the trivial subgroup? More generally, if $g\in G$, what about the set of powers of $g$? $\endgroup$ – Tobias Kildetoft Sep 9 '13 at 13:31
  • 2
    $\begingroup$ For extra credit: Show that every finite group $G$ is the union of its finite cyclic subgroups. Hint: If $x\in G$, can you think of a cyclic subgroup that contains $x$ and is finite? $\endgroup$ – Jyrki Lahtonen Sep 9 '13 at 13:36
  • $\begingroup$ take an element x, which is not identity, if its order is prime, done. suppose its order is $n = pq$, where p is prime and q is not 1. then what do you know about $x^q$? $\endgroup$ – Lost1 Sep 9 '13 at 14:13
  • $\begingroup$ @TobiasKildetoft Edit. Sorry, I forgot to mention that. $\endgroup$ – alop789312 Sep 9 '13 at 14:15

It's not necessary to invoke Cauchy here, because you are not asking for a specific-order cyclic subgroup. It is trivial that if $g \in G$ then $\langle g \rangle \leq G$ (which is the smallest subgroup which contains $g$).

And it's trivial that $\langle g \rangle \leq G$ is cyclic (in fact it contains elements of the form $g^m$ for some $m$.)

Then pick one non identity element of $G$ (it exists because $|G| \geq 2$), then it has order at least 2 (it contains at least $1_G$ and $g$) and so $\langle g \rangle \leq G$ is a cyclic subgroup of order at least 2.


Suppose that $G$ is a finite group of order $n$. Then there exists a prime $p$ with $p\mid n$. According to Cauchy's theorem, $G$ has an element $g$ of order $p$, and hence a cyclic subgroup $H=\langle g \rangle$ of order $p$. Cauchy's theorem can be seen as a partial converse to Lagrange's theorem. There is also a class of groups, so called CLT groups, for which the converse of Lagrange's theorem holds.

A finite group $G$ is a CLT-group if and only if for each positive divisor $d$ of $|G|$ there exists at least one subgroup $H\le G$ with $|H|=d$. It turns out that every CLT-group is solvable, and every supersolvable group is a CLT-group. For example, every finite group $G$ with $(G:Z(G))<12$ is a CLT-group. The result is best possible, because $A_4$ satisfies $(G:Z(G))=12$ and $A_4$ is not a CLT-group (it has no subgroup of order $6$, although $6\mid 12=|A_4|$).

  • 3
    $\begingroup$ bet you 100 quid the OP does not get a line of what you say in the 2nd paragraph $\endgroup$ – Lost1 Sep 9 '13 at 14:14
  • $\begingroup$ Cauchy's theorem is entirely unnecessary here, and I would guess unknown to the OP. $\endgroup$ – RghtHndSd Sep 9 '13 at 14:18
  • $\begingroup$ Yes, perhaps. But if you know already Lagrange's theorem, then Cauchy's theorem is not too far. And the question already mentiones a problem with the converse of Lagrange's theorem. $\endgroup$ – Dietrich Burde Sep 9 '13 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.