# Abstract Algebra: Every group has a cyclic subgroup

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 $$G$$.

• why such $k$ would exist?? Take $\mathbb{Z}$ under addition Jun 3, 2015 at 9:07
• Is that exactly how the question is stated? Jun 3, 2015 at 9:12
• @ drawnonward yes, the question is "Show that every group has a cyclic subgroup"
– Njal
Jun 3, 2015 at 9:13
• Trivially, I believe the identity element always serves as a cyclic subgroup. Jun 3, 2015 at 9:15
• @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. Jun 3, 2015 at 9:54

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.

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

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