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?
$\begingroup$
$\endgroup$
2
asked Mar 23, 2017 at 23:27
-
$\begingroup$ Isn't this true for any finite group? $\endgroup$– CrostulMar 23, 2017 at 23:38
-
1$\begingroup$ $\mathbb Z/4\mathbb Z$ is not generated by its element of order $2$. $\endgroup$– Misha LavrovMar 23, 2017 at 23:42
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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.
answered Mar 23, 2017 at 23:41
-
1$\begingroup$ Good argument! Indeed, it is reasonable that "the join of the atoms" is a characteristic subgroup of $G$, hence normal. $\endgroup$– CrostulMar 23, 2017 at 23:45
-
$\begingroup$ @Crostul What you say is obviously true, but the argument is exactly the same as Misha's! $\endgroup$ Mar 23, 2017 at 23:47