Why isn’t every subgroup of a Finite Group Cyclic?

I’m trying to prove or disprove wether every subgroup of a finite group is cyclic or not. I came up with this proof:

$$G$$ is a finite group of order $$k$$, with elements $$g_1,…,g_k$$, then every $$g_i\in G$$ will generate a cyclic subgroup $$H_i\subseteq G$$ such that $$H_i=\langle g_i \rangle$$, hence, $$G$$ has at most $$k$$ subgroups. The precise count of how many subgroups is not necessary. Finally, since every subgroup $$H_i=\langle g_i\rangle$$ is generated by a single element, every subgroup of $$G$$ is Cyclic.

Is this proof ok?

• So $G$ itself is cyclic ... Mar 19 at 8:24
• Why do you think that every subgroup is generated by only one element?
– Alex
Mar 19 at 8:33
• Why though? Isn’t the fact that $G$ is finite enough to ensure $\langle g \rangle$ will cycle back for every $g \in G$? Mar 19 at 8:34
• How about $\langle g_1, g_2 \rangle$?
– Alex
Mar 19 at 8:35
• Actually $\langle g\rangle$ is cyclic also when $G$ is infinite, by the way
– Alex
Mar 19 at 8:37

You have shown that every cyclic subgroup $$\langle g\rangle$$ of $$G$$ is cyclic. This is not very surprising. It doesn't imply that all subgroups of $$G$$ are cyclic, though. The easiest counterexample is $$G=S_n$$ with the subgroup $$H=A_n$$ for $$n\ge 4$$, which is not cyclic, because it is not even abelian.
• Your example is certainly correct but I would say that the easiest counterexample is $\mathbb{Z}_2\times\mathbb{Z}_2$ as a subgroup of $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$. Mar 19 at 12:41
• @AndreaMori Yes, I admit this is also easy. But the example $S_n$ and $A_n$ doesn't need three factors and also gives a nonabelian example. Mar 19 at 13:31
Take a non-cyclic group $$G$$ and any group $$H$$. Then, $$K:=G\times\{1_H\}\cong G$$ and $$K\le G\times H$$.