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Fermat's and Euler's Theorems
Lagrange's Theorem
Lagrange's Theorem
Proposition Let \(H\) be a subgroup of \(G\) with \(g \in G\) and define a map
\(\phi:H \rightarrow gH\) by \(\phi(h) = gh\).
The map \(\phi\) is bijective;
hence, the number of elements in \(H\) is the same as the number of elements in \(gH\).
Proof We first show that the map \(\phi\) is one-to-one.
Suppose that \(\phi(h_1) = \phi(h_2)\) for elements \(h_1, h_2 \in H\).
We must show that \(h_1 = h_2\),
but \(\phi(h_1) = gh_1\) and \(\phi(h_2) = gh_2\).
So \(gh_1 = gh_2\), and by left cancellation \(h_1= h_2\).
To show that \(\phi\) is onto is easy.
By definition every element of \(gH\) is of the form \(gh\) for some \(h \in H\) and \(\phi(h) = gh\).
Theorem: Lagrange Let \(G\) be a finite group and let \(H\) be a subgroup of \(G\).
Then \(|G|/|H| = [G : H]\) is the number of distinct left cosets of \(H\) in \(G\).
In particular,
the number of elements in \(H\) must divide the number of elements in \(G\).
Proof The group \(G\) is partitioned into \([G : H]\) distinct left cosets.
Each left coset has \(|H|\) elements;
therefore, \(|G| = [G : H] |H|\).
Corollary Suppose that \(G\) is a finite group and \(g \in G\).
Then the order of \(g\) must divide the number of elements in \(G\).
Corollary Let \(|G| = p\) with \(p\) a prime number.
Then \(G\) is cyclic and any \(g \in G\) such that \(g \neq e\) is a generator.
Proof Let \(g\) be in \(G\) such that \(g \neq e\).
Then by ,
the order of \(g\) must divide the order of the group.
Since \(|\langle g \rangle| \gt 1\), it must be \(p\).
Hence, \(g\) generates \(G\).
suggests that groups of prime order \(p\) must somehow look like \({\mathbb Z}_p\).
Corollary Let \(H\) and \(K\) be subgroups of a finite group \(G\) such that \(G \supset H \supset K\).
Then
\[\begin{aligned}\end{aligned}\].
Proof Observe that
\[\begin{aligned}\end{aligned}\].
In fact, we can say more about when two cycles have the same length.
Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications .
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x \in A element of, subset x belongs to A; every element of A is in B. \neq not equal The two sides are different. \sup supremum, infimum Least upper bound, greatest lower bound. \mathbb{N} number sets Naturals, integers, rationals, reals, complex numbers. \blacksquare\ \text{or}\ \square end of proof (halmos) Marks the point where the statement has been established. a \equiv b \pmod n congruent modulo n n divides a − b; a and b have the same remainder. a \mid b divides, greatest common divisor b is a multiple of a; the largest number dividing both. (G group, identity, inverse A set with an operation; the do-nothing element; the element that undoes g. G \cong H isomorphic, quotient group Same structure; the group of cosets of a normal subgroup N. \mathbb{Z}/n\mathbb{Z} integers modulo n The remainders 0…n−1 with clock arithmetic. \operatorname{Hom}(A arrows from A to B, composition The set of morphisms; do g then f.
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x \in A element of › \neq not equal › \sup supremum › \mathbb{N} number sets › \blacksquare\ \text{or}\ \square end of proof › a \equiv b \pmod n congruent modulo n › a \mid b divides › (G group › G \cong H isomorphic › \mathbb{Z}/n\mathbb{Z} integers modulo n › \operatorname{Hom}(A arrows from A to B ›
사람들이 묻는 질문
What is a group, in plain words? A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.
What is the difference between a ring and a field? A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.
이 페이지의 일부는 다음에서 변경되었습니다.
Judson, Abstract Algebra: Theory and Applications (GFDL 1.3).
여기서 압축하고 다시 설명; 오류는 우리의.
에 더 Abstract Algebra
Groups Subgroups, cosets and Lagrange's theorem Cyclic groups and permutation groups Homomorphisms, normal subgroups and quotient groups Rings and fields Galois theory: why the quintic has no formula