examples of infinite simple groups

Let X be a set and let f:XX be a function. Define


Throughout, we will say that f:XX is a permutationMathworldPlanetmath on X iff f is a bijection and C(f) is a finite setMathworldPlanetmath.

For permutation f:XX, the set C(f) will play the role of a ,,bridge” between the infiniteMathworldPlanetmath world and the finite world.

Let S(X) denote the group of all permutations on X (with compositionMathworldPlanetmath as a multiplication). For fS(X), subset AX will be called f-finite iff A is finite and C(f)A. This is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to the fact, that A is finite and if f(x)x, then xA.

It is easy to see, that if fS(X) and A is f-finite, then f(A)=A. Thus, we have well defined permutation (on a finite set) fA:AA by the formulaMathworldPlanetmathPlanetmath fA(x)=f(x).

Lemma. For any subset AX and any f,gS(X) such that A is f-finite and g-finite we have that A is fg-finite and


Proof. Assume, that A is f-finite and g-finite. Let xX be such that (fg)(x)x. Assume, that xA. Then f(x)=g(x)=x and thus (fg)(x)=x. ContradictionMathworldPlanetmathPlanetmath. Thus xA, so C(fg)A and since A is finite, then A is fg-finite. Finally, the equality


holds, because (fg)A is well definied (since A is fg-finite) and the operationMathworldPlanetmath ()A does not change the formulas of functions.

Now we can talk about the sign of a permutation. For fS(X) define


It can be easily checked, that sgn is well defined (indeed, sign depends only on those xX for which f(x)x). Furthermore, it follows directly from the definition, that


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


Briefly speaking, A(X) is the subgroupMathworldPlanetmathPlanetmath of even permutationsMathworldPlanetmath on a set X (a.k.a. the alternating groupMathworldPlanetmath for the set X).

Now, we shall prove the following propositionPlanetmathPlanetmath, 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 groupMathworldPlanetmathPlanetmath.

Proof. Assume, that A(X) is not simple and let NA(X) be a proper, nontrivial, normal subgroupMathworldPlanetmath. For a subset YX define

NY={fY|fN and C(f)Y}.

Note, that

A(Y)={fY|fA(X) and C(f)Y}.

Obviously NYA(Y) is a subgroup (due to lemma) of A(Y). We will show, that it is normal. Let fYNY and gYA(Y). We have to show, that gYfYgY-1NY. Of course


because N is normal (here f,g correspond to fY,gY). It follows from lemma (note, that Y is gfg-1-finite), that


which shows, that NY is normal. To obtain the contradiction, we need to show, that there exists YX with at least 5 elements, such that NY is nontrivial and proper (because in this case A(Y) is simple).

Let fN be such that fidX and let gA(X) be such that gN. Let Y be any f-finite and g-finite subset of X with at least 5 elements (such subset exists). Then NY is nontrivial, because fYNY is nontrivial.

Now assume, that gYNY, i.e. assume, that there exists hN with C(h)Y, such that gY=hY. Then (due to lemma) Y is hg-1-finite, and since gY=hY we have that for any xY the following holds:


On the other hand, for xX\Y we have g(x)=h(x)=x. This shows, that h=g, but hN and gN. Contradiction. Thus gYNY, so NY is proper.

This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

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

Title examples of infinite simple groupsPlanetmathPlanetmath
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