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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?

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  • $\begingroup$ 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 $d\ge 3$. $\endgroup$
    – YCor
    Commented Oct 14, 2018 at 22:57

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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.

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  • $\begingroup$ 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? $\endgroup$
    – RFZ
    Commented Oct 14, 2018 at 0:34
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    $\begingroup$ @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. $\endgroup$
    – Mark
    Commented Oct 14, 2018 at 5:27
  • $\begingroup$ 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. $\endgroup$
    – Hew Wolff
    Commented Oct 14, 2018 at 15:50
  • $\begingroup$ I will figure out also the case $d=0$. $\endgroup$ Commented Oct 14, 2018 at 15:52

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