# 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 ExamplesOfInfiniteSimpleGroups 2013-03-22 19:09:17 2013-03-22 19:09:17 joking (16130) joking (16130) 5 joking (16130) Example msc 20E32