1)state and prove largest theorem.
Answers
Answer:
Extension of Lagrange's theorem — If H is a subgroup of G and K is a subgroup of H, then
{\displaystyle [G:K]=[G:H]\,[H:K].}{\displaystyle [G:K]=[G:H]\,[H:K].}
Proof:
Let S be a set of coset representatives for K in H, so {\displaystyle H=\coprod _{s\in S}sK}{\displaystyle H=\coprod _{s\in S}sK} (disjoint union), and {\displaystyle |S|=[H:K]}{\displaystyle |S|=[H:K]}. For any {\displaystyle a\in G}{\displaystyle a\in G}, left-multiplication-by-a is a bijection {\displaystyle G\to G}G \to G, so {\displaystyle aH=\coprod _{s\in S}asK}{\displaystyle aH=\coprod _{s\in S}asK}. Thus each left coset of H decomposes into {\displaystyle [H:K]}{\displaystyle [H:K]} left cosets of K. Since G decomposes into {\displaystyle [G:H]}{\displaystyle [G:H]} left cosets of H, each of which decomposes into {\displaystyle [H:K]}{\displaystyle [H:K]} left cosets of K, the total number {\displaystyle [G:K]}{\displaystyle [G:K]} of left cosets of K in G is {\displaystyle [G:H][H:K]}{\displaystyle [G:H][H:K]}.
If we take K = {e} (e is the identity element of G), then [G : {e}] = |G| and [H : {e}] = |H|. Therefore we can recover the original equation |G| = [G : H] |H|.
A consequence of the theorem is that the order of any element a of a finite group (i.e. the smallest positive integer number k with ak = e, where e is the identity element of the group) divides the order of that group, since the order of a is equal to the order of the cyclic subgroup generated by a. If the group has n elements, it follows
{\displaystyle \displaystyle a^{n}=e{\mbox{.}}}\displaystyle a^{n}=e{\mbox{.}}
This can be used to prove Fermat's little theorem and its generalization, Euler's theorem. These special cases were known long before the general theorem was proved.
The theorem also shows that any group of prime order is cyclic and simple. This in turn can be used to prove Wilson's theorem, that if p is prime then p is a factor of {\displaystyle (p-1)!+1}(p-1)!+1.
Lagrange's theorem can also be used to show that there are infinitely many primes: if there were a largest prime p, then a prime divisor q of the Mersenne number {\displaystyle 2^{p}-1}2^{p}-1 would be such that the order of 2 in the multiplicative group {\displaystyle (\mathbb {Z} /q\mathbb {Z} )^{*}}{\displaystyle (\mathbb {Z} /q\mathbb {Z} )^{*}} (see modular arithmetic) divides the order of {\displaystyle (\mathbb {Z} /q\mathbb {Z} )^{*}}{\displaystyle (\mathbb {Z} /q\mathbb {Z} )^{*}}, which is {\displaystyle q-1}q-1. Hence p < q, contradicting the assumption that p is the largest prime.[2]
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