Definition of a group
If the non- empty set \(G\) and the algebraic structure \((G,⋅)\ ) formed by the binary operation \(⋅\) defined on \(G\ ) , satisfy:
- Closure: \(\forall a,b\in G\) , with \(a⋅b\in G\) .
- Associativity: \(\forall a,b,c\in G\) , with \((a⋅b)⋅c=a⋅(b⋅c)\) .
- Identity element: \(\exists e\in G\) , satisfying \(\forall a\in G\) with \(a⋅e=a\) .
- Inverse element: \(\forall a\in G\) , \(\exists b\in G\) make \(a⋅b=e\) , note \(a^{-1}=b\) .
Then the algebraic structure \((G,⋅)\) is a group .
Common groups are: integer, rational number, real number addition group; addition group in the sense of modulo \ (n\) ; multiplicative group composed of numbers that are relatively prime to \ (n\) in the sense of modulo\(n\); permutation The group, whose element is a bijection \(f\) , operates as a composition of maps.
Lagrange's theorem
For group \((G,⋅)\) , if there is \(G'\subset G\) and \((G',⋅)\) is also a group, then \((G',⋅)\) is A subgroup of \((G,⋅)\) , and \(|G|\) is a multiple of \(|G'|\) .
Proof:
Note that \(G_a\) represents the coset of the set \(G\) \(\{x⋅a|x\in G\}\) , then it is easy to know \(|G_a|=|G|\) .
For \(a,b\in G\) , if there is \(G'_a\cap G'_b \neq \emptyset\) , then \(\exists x,y\in G'\) satisfies \(x⋅ a=y⋅b ⇔ a=x^{-1}⋅y⋅b\) .
Then \(\forall z\in G'\) , there is \(z⋅a=z⋅(x^{-1}⋅y⋅b)=(z⋅x^{-1}⋅y)⋅b\ ) . It is easy to know \(z⋅x^{-1}⋅y \in G'\) , so every element in \(G'_a\) exists in \ (G'_b\) , that is \(G '_a=G'_b\) .
So it can be seen that there are only two relations between the cosets of \(G'\) , which are disjoint or identical. And since \(e\in G'\) , so \(G'\)The union of all cosets of is \(G\) . And since the size of the coset is equal to the original set, \(|G|\) is a multiple of \(|G'|\) .
Euler's theorem \(a^{\varphi(m)} \equiv 1 \pmod m\) can be derived from Lagrange's theorem .
Proof:
Set the set \(S=\{a_1,a_2,...,a_{\varphi(n)}\}\) , where \(gcd(a_i,n)=1\) . The algebraic structure formed by \(S\) and modular multiplication \((S,\times)\) is a group.
Then set \(S_i=\{1,a_i,a_i^2,a_i^3...\}\) , it is easy to know that \((S_i,\times)\) is \((S,\times)\) The subgroup of , namely \(|S_i||\varphi(n)\) . And \(a_i^{|S_i|}\equiv 1\) , so \(a_i^{\varphi(n)}\equiv 1\) .