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?
Best Answer
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.