1. From a category theoretical point of view every group has 1 element. In category theory we call an element of [math]A[/math] a morphism [math]g: 1 \rightarrow A [/math], where [math]1[/math] is a terminal object. Category of all groups [math]\mathsf{Grp}[/math] has 1 terminal object [math]\{ e \}[/math] - the trivial group. There is only 1 homomorphism to a group [math]G[/math] in a form [math]\{ e \} \rightarrow G[/math], the identity elements maps to the identity elements. Therefore, every group has 1 element, including finite groups.2. From a probabilistic point of view the set of all finite groups has probability 0. The set of all finite groups is countable, at most countable sets have lebesgue measure of 0. Therefore, set of all finite groups has probability 0.3. From a logical point of view finite theory of finite groups doesn't exist. By finite theory we mean that it can be generated by a finite number of axioms (a bit clickbait). Let's assume the opposite, we shall call the set of axioms [math]F[/math]. We are going to consider the union of [math]F[/math] and a set of propositions [math]\varphi_n[/math] for every n, that just states there exists at least n different elements. For any finite subset of this union there exists a model. Indeed, if it contains [math]\varphi_n[/math] for at most of n, then the model is [math]\mathbb{Z}_n[/math], if it only contains elements of [math]F[/math], then [math]\mathbb{Z}_n[/math] suffice too. Therefore, if we have a model for every finite subset of the union, and by compactness theorem the union has a model. However, for every n proposition [math]\varphi_n[/math] states that there are different n elements, the only model that it can have is an infinite group. Contradiction, because [math]F[/math] is axioms of finite groups.Finite groups are a psyop by the algebraic topologists establishment to not allow new people into algebraic topology. Anyway, this are 3 funny facts about finite groups, cheers.
>>167992842. Almost every finite group is trivial.
>>16799284>Set A (finite or infinite set) contains elements [a,b,c,...]>Set 1 contans the sets [A,B,C,...]>Set whatever contains the sets [1,2,3,...]By set theory you could preserve the rules of each subset or union the elements of each subset so your final set contains only the unique element list that is either finite or infinite.Overall it doesn't matter because your constructed sets exist for a purpose and you're just being a faggot.
>>16799284>its a scam because its too easywe know lol.
Damn it, I got nostalgic. https://www.youtube.com/watch?v=BipvGD-LCjU
>>16799772>be a finite simple group>SIMPLE>10 000 pages of classifiying them, 10 VOLUMES of pRoViNgI'm telling you it is all a psyop!!!
>>16799772Holy fuck the SOVL
>>16799284>From a category theoretical point of view every group has 1 elementNo, every group arises as the automorphisms of a single element.>The set of all finite groups is countable, at most countable sets have Lebesgue measure of 0.Yes, it is countable, but measure has to be defined in subsets of a measurable space. As far as I know, there is no measure over the class of all groups (up to isomorphism).Even if you drop a measure and just compare cardinalities, it is a "bad" criterion as we know that numbers which can be described by common language are countable, therefore you may conclude that they should be of no interest at all, which is weird as they are the most interesting/studied of all! Perhaps, cardinalities just confirm that humans are biased, and love particulars instead of generals.At last, for your attempt of proving that the axioms of finite groups do not make sense, you make a very basic logic error. The correct set of axioms is to let [math]\varphi_n[/math] be the axiom of the group having at most [math]n[/math] elements, and the union would be the theory of finite groups, which do admit models.
>>16801252>No, every group arises as the automorphisms of a single elementI didn't get your argument. In category theory we call an element of an object [math]A[/math] a morphism from the terminal object [math]1[/math] to [math]A[/math]. We don't care about morphisms [math]A \rightarrow A[/math] (automorphisms are just like that). In [math]\mathsf{Grp}[/math] there is only one morphism of the form [math]1 \rightarrow G[/math] for any group [math]G[/math] ([math]e \mapsto e_G[/math]).>has to be defined in subsets of a measurable spaceBy Cayley's theorem we can embed every finite group to a symmetric group (its subgroup). On [math]\mathbb{R}[/math] there is obviously a defined measure.>your third case is wrongHow it is? I will explain with details. Let's consider a finite subset of [math]F \cup {\varphi_n}[/math]. If this finite subset contains only elements of [math]F[/math] then any finite group will suffice. If it contains elements of [math]F[/math] and some elements of the form [math]\varphi_n[/math] then any finite group that has [math]m[/math] elements will suffice, where [math]m[/math] is such a number that [math]\varphi_k[/math] with [math]k > m[/math] isn't contained in our finite subset. If it contains only elements of the form [math]\varphi_n[/math] then any finite group that has [math]m[/math] elements will suffice, where [math]m[/math] is defined as in the previous case. For any finite subset we have a model, therefore by compactness theorem [math]F \cup {\varphi_n}[/math] has a model. However, such a model must be a finite group because of [math]F[/math] but must have 1, 2, 3, ... different elements. How this can be, if a finite group is finite? Only infinite groups have infinite different elements. Thus a contradiction.P.S: Thanks for your post! If you think there is still a mistake, I will gladly answer. Maybe I did a mistake after all, although this thread I made just as a joke.
