Prove that every finite group is generated by its cyclic subgroups.

I'm trying to prove that every finite group is generated by its cyclic subgroups.

Let $$|G|=n$$ and $$G=\langle g_1,\ldots,g_n\rangle$$. Then $$G=\langle g_1,\ldots,g_n\rangle\leq \bigl\langle\langle g_1\rangle,\ldots,\langle g_n\rangle\bigr\rangle\leq G$$, since every $$\langle g_i\rangle \leq G$$. Hence $$G=\bigl\langle\langle g_1\rangle ,\ldots,\langle g_n\rangle\bigr\rangle$$.

That's the right way to solve this problem? if it's not, how can I show it?

• What do you mean by $G$ being generated by its cyclic subgroups? That $G$ is the union of all those subgroups? Commented Oct 3, 2021 at 15:59
• every group is generated by its cyclic subgroups, right? it's generated by its set of elements, and any element is contained in the cyclic subgroup it generates. Commented Oct 3, 2021 at 16:09
• Why are you assuming that $G$ is finite? It seems clear that every group, finite or infinite, is generated by its cyclic subgroups, because, if $H$ is the subgroup of $G$ generated by the cyclic subgroups of the group $G$, then for all $g \in G$ we have $g \in \langle g \rangle \le H$, so $H=G$. Commented Oct 3, 2021 at 16:16
• In fact, every group is the union of its cyclic groups
– lhf
Commented Oct 3, 2021 at 17:29

Yes, your reasoning is correct and it generalizes to the case where $$G$$ is not necessarily finite:
$$G = \langle g : g \in G \rangle \subseteq \langle \langle g \rangle : g \in G \rangle \subseteq G.$$