# Does every finite group have a finite cyclic subgroup?

Does every finite group of size at least 2 have a 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.

• What about the trivial subgroup? More generally, if $g\in G$, what about the set of powers of $g$? Commented Sep 9, 2013 at 13:31
• 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? Commented Sep 9, 2013 at 13:36
• 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 $x^q$? Commented Sep 9, 2013 at 14:13
• @TobiasKildetoft Edit. Sorry, I forgot to mention that. Commented Sep 9, 2013 at 14:15
• Lagrange's theorem actually does have a partial (true) converse: if $p$ is a prime dividing the order of $G$, then there is an element $g\in G$ of order $p$. Commented Jul 2, 2020 at 21:49

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 $6\mid 12=|A_4|$).

• bet you 100 quid the OP does not get a line of what you say in the 2nd paragraph Commented Sep 9, 2013 at 14:14
• Cauchy's theorem is entirely unnecessary here, and I would guess unknown to the OP. Commented Sep 9, 2013 at 14:18
• 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. Commented Sep 9, 2013 at 14:23