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?

  • 1
    $\begingroup$ So $G$ itself is cyclic ... $\endgroup$
    – Paul Frost
    Mar 19 at 8:24
  • 1
    $\begingroup$ Why do you think that every subgroup is generated by only one element? $\endgroup$
    – Alex
    Mar 19 at 8:33
  • $\begingroup$ Why though? Isn’t the fact that $G$ is finite enough to ensure $\langle g \rangle$ will cycle back for every $g \in G$? $\endgroup$ Mar 19 at 8:34
  • $\begingroup$ How about $\langle g_1, g_2 \rangle$? $\endgroup$
    – Alex
    Mar 19 at 8:35
  • $\begingroup$ Actually $\langle g\rangle$ is cyclic also when $G$ is infinite, by the way $\endgroup$
    – Alex
    Mar 19 at 8:37

2 Answers 2


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.

  • 2
    $\begingroup$ 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 $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$. $\endgroup$ Mar 19 at 12:41
  • $\begingroup$ @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. $\endgroup$ Mar 19 at 13:31
  • $\begingroup$ Convincing as usual. $\endgroup$
    – Mikasa
    Mar 19 at 17:46

Take a non-cyclic group $G$ and any group $H$. Then, $K:=G\times\{1_H\}\cong G$ and $K\le G\times H$.


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