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Let $G$ be a simple group and there exists a subgroup $H$ of index $n \in \mathbb{N}_{\ge 3}$. Show that: $|G|$ divides $n!/2.$ I saw a similar question here.

There was written: "The proof is really simple. Let $X$ be the set of left cosets of $H$. Consider $\phi: G \to \text{Sym}(X)$ given by $\phi(x)(aH)=(xa)H$. Then $\phi$ is a homomorphism. Consider now $K=\ker \phi$. Then $K$ is a normal subgroup of $G$ contained in $H$. Finally, $G/K$ is isomorphic to a subgroup of $\text{Sym}(X)$, which has order $n!$, where $n=[G:H]$. Thus, $[G:K]$ is finite and divides $n!$."

In fact, I don't understand the last sentence. "Thus, $[G:K]$ is finite and divides $n!$." Also did I understand it right that $ker(\phi)=H$? Can you also give me an advice how to proof it for $n \ge 3$. (maybe induction)

Thank you for taking your time.

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maiT
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1 Answers1

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Sym(X), is the group of permutations of the collections of cosets, $G/H$. As $H$ has index $n$, this group Sym($X$) is of order $n!$. As $K$ is the kernal of $\phi$ we see$\phi$ leads to an injective homomorphism $G/K\to Sym(X)$ and by Lagrange's theorem its order should divide $n!$. As there is anormal subgroup of index 2 (consisting of even permutations) in Sym($X$), the image of the simple group $G$ should be contained in that normal subgroup otherwise its intersection will lead to a proper normal subgroup in $G$, contradicting its simplicity. So $|G|$ divides $n!/2$

EDIT: I should explore if it is possible for image of $\phi$ to have trivial intersection with Alt(X). In that case the sign homomorphism $S_n\to\{\pm1\}$ restricts to a non-trivial homomorphism on Image of $G$, whose kernel gives a proper nontrivial normal subgroup in image of $G$ and hence on $G$ contradicting the simplicity of $G$.

P Vanchinathan
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  • Why there is a subgroup of index 2? – maiT Nov 01 '16 at 09:11
  • Alt(X) is a subgroup of index 2 in Sym(X), that is of order $n!/2$. – P Vanchinathan Nov 01 '16 at 09:14
  • But $S_X$ isn't the image of $G$. – Stefan4024 Nov 01 '16 at 09:15
  • Doesn't have to be. The image is a subgroup of Alt(X). – P Vanchinathan Nov 01 '16 at 09:16
  • First of all you use explicitly say that "$S_x$, the image of the simple group $G$". Second of all you should explain why the image of $G$ is a subgroup of $A_X$, which is true, but not obvious. – Stefan4024 Nov 01 '16 at 09:18
  • Read the last but one statement in my answer. – P Vanchinathan Nov 01 '16 at 09:20
  • In my opinion you still haven't discarded the possibility of a trivial intersection, i.e. $\phi[G] \cap A_5 = {e}$, which is possible, as the fact that $G$ is simple isn't violated. – Stefan4024 Nov 01 '16 at 09:25
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    I'll edit the answer now to rule out this possibility. – P Vanchinathan Nov 01 '16 at 09:27
  • sry can you explain the last step to me once more, I try to understand it but it doesn't work ... – maiT Nov 01 '16 at 10:02
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    @milui If $\sigma: S \to {\pm 1}$, then if the intersection of $\phi[G]$ and $A_5$ is trivial then we must have both odd and even permutations in $\phi[G]$, so then $\phi \sigma$ is a surjective homomorphism and the kernel of $\phi \sigma$ is a proper normal subgroup of $G$, which is impossible.

    BTW Vanchinathan, now it looks better.

     – Stefan4024 Nov 01 '16 at 10:14