# Is a finite simple group generated by its minimal subgroups?

The question in title can be reformulated as follows:

Is a finite simple group the join of the atoms of its subgroup lattice?

or as follows:

Can a finite simple group be generated by a set of elements of prime order?

• Isn't this true for any finite group? – Crostul Mar 23 '17 at 23:38
• $\mathbb Z/4\mathbb Z$ is not generated by its element of order $2$. – Misha Lavrov Mar 23 '17 at 23:42

## 1 Answer

If $G$ is not generated by all its elements of prime order, let $H$ be the subgroup they do generate. I claim that $H$ is normal, and therefore $G$ is not simple.

Given any $h \in H$, we can write $h = h_1 h_2 \dotsb h_k$, where each $h_i$ is an element of $H$ with prime order. For an arbitrary $g \in G$, we have $$ghg^{-1} = (gh_1g^{-1})(gh_2g^{-1})\dotsb(gh_kg^{-1})$$ and if $(h_i)^{p_i} = e$ then $(gh_ig^{-1})^{p_i} = e$, so this is also a product of elements with prime order. Therefore $ghg^{-1} \in H$, proving that $H$ is normal.

• Good argument! Indeed, it is reasonable that "the join of the atoms" is a characteristic subgroup of $G$, hence normal. – Crostul Mar 23 '17 at 23:45
• @Crostul What you say is obviously true, but the argument is exactly the same as Misha's! – Igor Rivin Mar 23 '17 at 23:47