# examples of infinite simple groups

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

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

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 $f(x)\ne x$, 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 $(f\circ g)(x)\ne x$. Assume, that $x\notin A$. 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. $\mathrm{\square}$

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 $f(x)\ne x$). 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

$${N}_{Y}=\{{f}_{Y}|f\in N\text{and}C(f)\subseteq Y\}.$$ |

Note, that

$$A(Y)=\{{f}_{Y}|f\in A(X)\text{and}C(f)\subseteq Y\}.$$ |

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

$${g}_{Y}\circ {f}_{Y}\circ {g}_{Y}^{-1}={(g\circ f\circ {g}^{-1})}_{Y}\in {N}_{Y},$$ |

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 $f\ne {\mathrm{id}}_{X}$ and let $g\in A(X)$ be such that $g\notin N$. 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 $g\notin N$. Contradiction. Thus ${g}_{Y}\notin {N}_{Y}$, so ${N}_{Y}$ is proper.

This completes^{} the proof. $\mathrm{\square}$

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 |