# Problem about index of proper nontrivial subgroup

Show that a finite simple group $$G$$ of order $$\geq d!$$ can not have a proper nontrivial subgroup of index $$d$$.

Remark: I guess that the condition of this problem is a bit incorrect, namely we need to put $$|G|>d!$$ instead of $$|G|\geq d!$$.

Proof: Suppose $$\exists H \lneq G$$, $$H\neq \{e\}$$ such that $$[G:H]=d$$. Consider the set $$S=\{xH: x\in G\}$$ the set of all left cosets of $$H$$ in $$G$$. Consider the action of $$G$$ on a set $$S$$ by left multiplication then we get homomorphism $$\phi:G\to \text{Sym}(S)$$ by $$\phi(g)=\pi_g$$ where $$\pi_g:S\to S$$ defined by $$\pi_g(xH)=gxH$$. It is not so hard to check that $$\text{ker} \phi=\bigcap \limits_{x\in G}xHx^{-1}$$ and $$\text{ker} \phi$$ is normal and is the largest normal subgroup of $$G$$ contained in $$H$$. Since we supposed that $$H\lneq G$$ and $$G$$ is simple then $$\text{ker} \phi=\{e\}$$ and hence $$\phi$$ is injective. Hence $$\phi(G)\subseteq \text{Sym}(S)$$ so $$|\phi(G)|=|G|\leq |\text{Sym}(S)|$$.

In order to get contradiction we need assume that $$|G|>d!$$ because in this case we'll get $$d!<|G|\leq |\text{Sym}(S)|=d!$$. Am i right?

• No, it's correct (for $d\ge 3$). Actually $\ge d!$ can be replaced with $>\max(d!/2,2)$. Indeed, this implies that the signature map is trivial, and then get a homomorphism into the alternating group, which has to be trivial, and hence the subgroup is trivial, so $G$ has order $d$, and $d!/2\ge d$ for $d\ge 3$. – YCor Oct 14 '18 at 22:57

But if $$|G|=d!$$ there are two possibilities. 1. $$\phi$$ is not injective. Done since the kernel is not trivial and normal. If $$\phi$$ is injective, it is an isomorphism contradiction since $$S_d=Sym(S)$$ is not simple.
• I don't understand your answer. Why are you considering the case when $\phi$ is not injective? the kernel of homomorphism is trivial then mapping is injective. Could you clarify your answer? – ZFR Oct 14 '18 at 0:34
• @RFZ, you are right that it is easy to see $Ker(\phi)\ne G$ so it must be $Ker(\phi)=\{e\}$ and hence $\phi$ is injective. You got the contradiction right if $|G|>d!$. But Tsemo Aristide tries to explain that you also need to find a contradiction in the case when $|G|=d!$. So suppose $|G|=d!$ and $\phi$ is injective. Then you get that $G$ is isomorphic to $S_d$ and that means $S_d$ is simple. But that is a contradiction because we know $S_d$ is not simple, $A_d$ is its non trivial normal subgroup. – Mark Oct 14 '18 at 5:27
• Finally we should mention the case $d = 1$, which is not covered by the above argument (because $S_1$ is simple) but is trivial: a subgroup of index $1$ cannot be proper. – Hew Wolff Oct 14 '18 at 15:50
• I will figure out also the case $d=0$. – Tsemo Aristide Oct 14 '18 at 15:52