# If $G$ is an inﬁnite simple group then any proper subgroup of $G$ has inﬁnite index.

If $$G$$ is an inﬁnite simple group then any proper subgroup of $$G$$ has inﬁnite index.

This question's hint is use the $$n!$$-theorem but i dont understand how i use it for answer.

$$n!$$-theorem: Let $$G$$ be a group and $$H$$ be a subgroup of $$G$$ of finite index, say $$|G:H|=n$$. Then there is a normal subgroup N of $$G$$ such that $$N\subseteq H$$ and $$G/N$$ is isomorphic to a subgroup of $$S_n$$ and so $$|G/N|$$ divides $$n!$$. Indeed, $${\rm core}_G(H)$$ is such a normal subgroup of $$G$$.

• How did this problem arise? – Shaun Oct 29 '19 at 17:58
• You are misquoting the theorem. The theorem guarantees that $N\subseteq H$, not merely $N\subseteq G$. – Arturo Magidin Oct 29 '19 at 18:21
• i edited, thank you – user709858 Oct 29 '19 at 18:23
• The existence of a normal subgroup of finite index contradicts $G$ being an infinite simple group. – Derek Holt Oct 29 '19 at 18:34
• No, I have just told you how to prove the result. It is easy. – Derek Holt Oct 29 '19 at 18:48

Assume for contradiction that $$G$$ is an infinite simple group and has a finite index subgroup $$H$$ such that $$|G:H|>1$$. Then, $$\mathrm{Core}_G(H)$$ is a finite index normal subgroup because the intersection of finite index subgroups is finite and by definition $$\mathrm{Core}_G(H)$$ is invariant under conjugation by $$G$$. In particular, $$G$$ has a non-trivial proper normal subgroup and therefore cannot be simple. Thus, $$G$$ does not have any proper finite index subgroups.
Now given a finite index subgroup $$H$$, it has only finitely many conjugates by 1. So intersect all of them. This subgroup has finite index by 2. It is clearly normal because conjugating the intersection amounts to intersecting the conjugates, but that just permutes the things we're intersecting and we end up intersecting the same finite set of subgroups.