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I would be very thankful if someone could give me a hint with proving next statement:

If $G$ is a group with a subgroup $H$ of finite index $n$, then $G$ has a normal subgroup $K$ contained in $H$ whose index in $G$ is finite and divides $n!$.

I found the proof on this page: Wikipedia page on index of a subgroup, but I get lost in the part of it which proves that the sets $A$ and $B$ have the same cardinality.

Thanks!!

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2 Answers 2

up vote 10 down vote accepted

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 $n!$.

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Let $G$ be a finitely generated group.

An alternative result is the following:

Thm: There exists a characteristic subgroup of index $(n!)^{2n}$ in $G$.

A characteristic subgroup is one which is fixed (not necessarily pointwise) by all automorphisms of $G$.

Firstly, note that there are only $(n!)^n$ many subgroups of index $n<\infty$ in your group. A proof of this can be found here, and is somewhat similar to lhf's answer (I'm sure this bound is much tighter!).

Then, if you intersect two groups of finite index, $H$ and $K$ say, you get another group of finite index, $H\cap K$, and $|G:H\cap K|\leq |G:H|\cdot |G:K|$. To see this, let $L=H\cap K$ and let $a\in Lb$ where $Lb$ is one of your finite number of cosets. Then $ab^{-1}\in L\Rightarrow a\in Hb\cap Kb\Rightarrow Lb\leq Hb\cap Kb$. Clearly $Lb\leq Hb\cap Kb$, so $Hb\cap Lb=Db$. Thus, the number of cosets of $L$ is $\leq |G:H|\cdot |G:K|$ as required.

So intersecting all subgroups of a fixed index $n$ will give you a subgroup of order $(n!)^{2n}$ (so, slightly weaker). Moreover, this subgroup is characteristic. It is characteristic because automorphisms fix the index of subgroups. That is, if $\phi$ is an automorphism of $G$ then $|G:H|=|G:H\phi|$. So, $\cap (H_i\phi)=\cap H_i$ so your group is fixed by all automorphisms.

Application:

The following application of the above is a theorem of Gilbert Baumslag from 1963. It is quite unusual, as it is surprising (residual finiteness is usually very hard to prove), and it has a very short proof (the paper is a page-and-a-half long, and contains three applications of this result. The half-page is all references.)

Thm: If $G$ is a finitely generated residually finite group then $\operatorname{Aut}(G)$ is residually finite.

A group is residually finite if for any element $g\in G$ there exists a homomorphism onto a finite group $F$, $\phi: G\rightarrow F$ say, such that $g\phi\neq 1$. This is a very strong finiteness condition. Equivalently, all the finite index subgroups intersect trivially (we proved above that finitely many intersect with finite index, but there are infinitely many in general so this makes sense).

Proof: Let $1\neq \alpha\in \operatorname{Aut(G)}$. Then there exists $g\in G$ such that $g\alpha\neq g$ and write $h=(g\alpha)g^{-1} (\neq 1)$. As the finite index subgroups intersect non-trivially, there is a finite index subgroup of $G$ not containing $h$, $K$ say. One can take this to be characteristic, by intersecting all subgroups of that index, as above (if $h\not\in A$ then $h\not\in A\cap B$ whatever $B$ is). Now, $\operatorname{Aut(G)}$ induces a finite group ($A\cong \operatorname{Aut(G)}/L$) of automorphism of $G/K$, as $G/K$ if finite and $K$ is characteristic. As $h\not\in K$ we have that $\alpha$ induces a non-trivial automorphism of $G/K$, so $\alpha\not\in L$, and so $\operatorname{Aut(G)}$ is residually finite, as required.

Grossman extended this result to $\operatorname{Out}(G)$ ("On the residual finiteness of certain mapping class groups", 1974).

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Tis proof is still alive after one year. +1 –  B. S. Jun 26 '13 at 17:04

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