# Subgroups of finite simple groups.

Let $$G$$ be a nontrivial finite group and $$X_G$$ the set of all the proper subgroups of $$G$$, $$X_G:=\{H\subseteq G\mid H\le G \wedge H\ne G\}$$. Lagrange's theorem put a limitation on the order of the elements of $$X_G$$, which must be a proper divisor of $$|G|$$. If $$G$$ is simple, then an additional limitation on the order of the elements of $$X_G$$ gets in, namely:

$$G\space\text{simple}\Longrightarrow[G:H]!\ge|G|, \space\forall H\in X_G \tag 1$$

In fact, for $$H\in X_G$$, the group $$G$$ acts by left multiplication on the left quotient set $$G/H$$, and this action has trivial kernel$$^\dagger$$. So, $$G$$ embeds into $$S_{[G:H]}$$, whence $$(1)$$.

The constraint $$(1)$$ means that finite simple groups can't have "relatively big" subgroups. For example, from $$(1)$$ follows that $$A_5$$ can't have subgroups of order $$15$$, $$20$$ and $$30$$ (this latter is ruled out by the very simplicity of $$A_5$$, though), because $$[60:k]!<60$$ for $$k=15, 20, 30$$.

Does the inverse implication in $$(1)$$ hold, too? Some thoughts. A counterexample would be a nonsimple finite group with all its proper subgroups fulfilling $$(1)$$. $$|G|=2,3$$: it is $$G\cong C_2,C_3$$, both simple, so we have to move upwards. $$|G|=4$$: both classes ($$C_2\times C_2$$ and $$C_4$$) are nonsimple, but both have subgroups of index $$2$$ (and $$2!<4$$). $$|G|=5$$: $$C_5$$ is simple. $$|G|=6$$: likewise $$|G|=4$$ case, both classes ($$S_3$$ and $$C_6$$) are nonsimple, but both have subgroups of index $$2$$ (and $$2!<6$$). $$|G|=7$$: $$C_7$$ is simple. Therefore, counterexamples (if any) must have $$|G|\ge 8$$.

$$^\dagger$$For $$H\in X$$, we get: $$K:=\bigcap_{g\in G}gHg^{-1}\lneq G$$, and thence $$K=\{1\}$$ for the simplicity of $$G$$.

• Isn't ${\rm SL}(2,5)$ a counterxample? The largest proper subgroup has index $5$, so you get equality on the RHS of (1). Commented Feb 27, 2021 at 10:26

Let $$p$$ be a prime and $$k > 1$$ such that $$p^k \leq p!$$. Then any finite group $$G$$ of order $$p^k$$ satisfies your condition, since any subgroup of $$G$$ has index $$\geq p$$. (For example consider $$G$$ elementary abelian of order $$25 = 5^2$$.)
You could also look any group of order $$n = p_1^{k_1} \cdots p_t^{k_t}$$, where $$p_i$$ are distinct primes such that $$n \leq p_i!$$ for all $$1 \leq i \leq t$$.