examples of infinite simple groups

Let $X$ be a set and let $f:X\to X$ be a function. Define

C(f)=\{x\in X|f(x)\neq x\}.

Throughout, we will say that $f:X\to X$ is a permutation on $X$ iff $f$ is a bijection and $C(f)$ is a finite set.

For permutation $f:X\to X$ , the set $C(f)$ will play the role of a ,,bridge” between the infinite world and the finite world.

Let $S(X)$ denote the group of all permutations on $X$ (with composition as a multiplication). For $f\in S(X)$ , subset $A\subset X$ will be called $f$ -finite iff $A$ is finite and $C(f)\subseteq A$ . This is equivalent to the fact, that $A$ is finite and if $f(x)\neq x$ , then $x\in A$ .

It is easy to see, that if $f\in S(X)$ and $A$ is $f$ -finite, then $f(A)=A$ . Thus, we have well defined permutation (on a finite set) $f_{A}:A\to A$ by the formula $f_{A}(x)=f(x)$ .

Lemma. For any subset $A\subseteq X$ and any $f,g\in S(X)$ such that $A$ is $f$ -finite and $g$ -finite we have that $A$ is $f\circ g$ -finite and

(f\circ g)_{A}=f_{A}\circ g_{A}.

Proof. Assume, that $A$ is $f$ -finite and $g$ -finite. Let $x\in X$ be such that $(f\circ g)(x)\neq x$ . Assume, that $x\not\in A$ . Then $f(x)=g(x)=x$ and thus $(f\circ g)(x)=x$ . Contradiction. Thus $x\in A$ , so $C(f\circ g)\subseteq A$ and since $A$ is finite, then $A$ is $f\circ g$ -finite. Finally, the equality

(f\circ g)_{A}=f_{A}\circ g_{A}

holds, because $(f\circ g)_{A}$ is well definied (since $A$ is $f\circ g$ -finite) and the operation $(\cdot)_{A}$ does not change the formulas of functions. $\square$

Now we can talk about the sign of a permutation. For $f\in S(X)$ define

\mathrm{sgn}(f)=\mathrm{sgn}(f_{A}).

It can be easily checked, that $\mathrm{sgn}$ is well defined (indeed, sign depends only on those $x\in X$ for which $f(x)\neq x$ ). Furthermore, it follows directly from the definition, that

\mathrm{sgn}:S(X)\to\{-1,1\}

is a group homomorphism (in $\{-1,1\}$ we have standard multiplication). Define

A(X)=\mathrm{ker}(\mathrm{sgn}).

Briefly speaking, $A(X)$ is the subgroup of even permutations on a set $X$ (a.k.a. the alternating group for the set $X$ ).

Now, we shall prove the following proposition, using the fact, that for any finite set $X$ with at least $5$ elements, the group $A(X)$ is simple (this is well known fact).

Proposition. If $X$ is an infinite set, then $A(X)$ is a simple group.

Proof. Assume, that $A(X)$ is not simple and let $N\subseteq A(X)$ be a proper, nontrivial, normal subgroup. For a subset $Y\subseteq X$ define

N_{Y}=\{f_{Y}\ |\ f\in N\mbox{ and }C(f)\subseteq Y\}.

Note, that

A(Y)=\{f_{Y}\ |\ f\in A(X)\mbox{ and }C(f)\subseteq Y\}.

Obviously $N_{Y}\subseteq A(Y)$ is a subgroup (due to lemma) of $A(Y)$ . We will show, that it is normal. Let $f_{Y}\in N_{Y}$ and $g_{Y}\in A(Y)$ . We have to show, that $g_{Y}\circ f_{Y}\circ g_{Y}^{-1}\in N_{Y}$ . Of course

g\circ f\circ g^{-1}\in N,

because $N$ is normal (here $f, g$ correspond to $f_{Y},g_{Y}$ ). It follows from lemma (note, that $Y$ is $g\circ f\circ g^{-1}$ -finite), that

g_{Y}\circ f_{Y}\circ g^{-1}_{Y}=(g\circ f\circ g^{-1})_{Y}\in N_{Y},

which shows, that $N_{Y}$ is normal. To obtain the contradiction, we need to show, that there exists $Y\subseteq X$ with at least $5$ elements, such that $N_{Y}$ is nontrivial and proper (because in this case $A(Y)$ is simple).

Let $f\in N$ be such that $f\neq\mathrm{id}_{X}$ and let $g\in A(X)$ be such that $g\not\in N$ . Let $Y$ be any $f$ -finite and $g$ -finite subset of $X$ with at least $5$ elements (such subset exists). Then $N_{Y}$ is nontrivial, because $f_{Y}\in N_{Y}$ is nontrivial.

Now assume, that $g_{Y}\in N_{Y}$ , i.e. assume, that there exists $h\in N$ with $C(h)\subseteq Y$ , such that $g_{Y}=h_{Y}$ . Then (due to lemma) $Y$ is $h\circ g^{-1}$ -finite, and since $g_{Y}=h_{Y}$ we have that for any $x\in Y$ the following holds:

(h\circ g^{-1})(x)=x.

On the other hand, for $x\in X\backslash Y$ we have $g(x)=h(x)=x$ . This shows, that $h=g$ , but $h\in N$ and $g\not\in N$ . Contradiction. Thus $g_{Y}\not\in N_{Y}$ , so $N_{Y}$ is proper.

This completes the proof. $\square$

Remark. This proposition shows, that the class of simple groups is actually a proper class, i.e. it is not a set. Therefore studying infinite simple groups can be very difficult.

Title	examples of infinite simple groups
Canonical name	ExamplesOfInfiniteSimpleGroups
Date of creation	2013-03-22 19:09:17
Last modified on	2013-03-22 19:09:17
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	5
Author	joking (16130)
Entry type	Example
Classification	msc 20E32