I have to show that every group has a cyclic subgroup. I know what this means, and to me it is obvious, yet I am not sure how to formally write it.

I proved it directly, as follows:

Let $G$ be a group.

Let $g$ be an element in $G$.

Let $o(g) = k$, i.e., $g^k = e$, for some integer $k$,

so by definition, $\langle g\rangle = \{e, g, g^2, \dots, g^{k-1}\}$,

therefore, for some $g\in G$, the cyclic subgroup generated by $g$ is in fact a subgroup of $G$.

Would that be sufficient to prove the statement? Did I leave anything out, or should I mention anything else?

New version of the proof:

Let $G$ be a finite group.

If $G = \{e\}$, then $G = \langle e\rangle$, i.e., cyclic.

If $G\neq\{e\}$, then there exists $g\neq e$, for $g\in G$.

Let $m$ be minimum positive integer s.t. $g^m = e$.

Then $g, g^2, \dots, g^m = e$ are distinct elements in $G$ generated by $g$

So $e, g, g^2, \dots, g^{m-1}$ form a cyclic subgroup $\langle g\rangle$ in $G$.

  • 1
    $\begingroup$ why such $k$ would exist?? Take $\mathbb{Z}$ under addition $\endgroup$ Jun 3, 2015 at 9:07
  • $\begingroup$ Is that exactly how the question is stated? $\endgroup$ Jun 3, 2015 at 9:12
  • $\begingroup$ @ drawnonward yes, the question is "Show that every group has a cyclic subgroup" $\endgroup$
    – Njal
    Jun 3, 2015 at 9:13
  • 4
    $\begingroup$ Trivially, I believe the identity element always serves as a cyclic subgroup. $\endgroup$ Jun 3, 2015 at 9:15
  • 1
    $\begingroup$ @kritzikratzi , I don't think it is cheating, because in some groups, it will be the only proper cyclic subgroup....for example $\mathbb{Z}/p\mathbb{Z}$ I think the purpose here is to assume the group has finite cardinality = n, then n=pm, p prime and by lagranges theorem we are done. $\endgroup$ Jun 3, 2015 at 9:54

2 Answers 2


Your proof is incorrect. You take an element $g\in G$ and immediatelly claim that $o(g) = k$ for some integer $k$. However, you cannot claim that, as there are groups in which no element has finite order.

  • $\begingroup$ Should I first prove that the cyclic subgroup has order o(g)? $\endgroup$
    – Njal
    Jun 3, 2015 at 9:13
  • $\begingroup$ @Njal Hint $\mathbb Z$ is also a cyclic group. $\endgroup$
    – 5xum
    Jun 3, 2015 at 9:16
  • $\begingroup$ Ok, I think I know what to do, going to edit the post $\endgroup$
    – Njal
    Jun 3, 2015 at 12:09

As already said, it is not true that an arbitrary group has always a finite cyclic subgroup: take $\Bbb Z$ under addition.

However, what is true, is that an arbitrary group $G$ has always a cyclic subgroup. Take an element $g\in G$ and consider the subgroup of $G$ generated by this element: $\langle g\rangle$.

You have now two cases:

1)$\operatorname{ord}(g)$ is finite, say $N$; then $$ \langle g\rangle=\{1_G,g,g^2,\dots,g^{N-1}\}=\{g^k\;:\;0\le k\le N-1\} $$

for example you can take $G=S_n$ the symmetric group and $g=(1,2,\dots,n)$, or even $G=\Bbb Z_n$ the class of remainder modulo $n$, which is a group under addition, and take $g=1$. An interesting example is the one given by the group $GL_n(\Bbb R)$ which is an infinite group taking a idempotent matrix as $g$.

2)$\operatorname{ord}(g)=\infty$: the situation is as above, except the fact that the list of elements in $\langle g\rangle$ never ends, i.e. $$ \langle g\rangle=\{1_G,g,g^2,\dots,g^{k},\dots\}=\{g^k\;:\;k\in\Bbb Z_{\ge 0}\} $$ for example take $G=\Bbb Z$ with $g=1$ or even $GL_n(\Bbb R)$ as above with a non idempotent matrix as $g$.

  • $\begingroup$ This is a wrong answer. Every group has a finite cyclic subgroup $\{1\}$. $\endgroup$
    – markvs
    Jul 2, 2020 at 20:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .