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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 .
ປັດຈຸບັນທ່ານ
ບໍ່ມີເຄື່ອງຄິດໄລ່ທີ່ຈະແກ້ໄຂບັນຫານີ້ໄດ້, ແຕ່ສ່ວນຂອງມັນສາມາດຄິດໄລ່ໄດ້. ພະຍາຍາມອັນໃດອັນໜຶ່ງຂ້າງລຸ່ມນີ້ ຫຼື ພິມຕົວເອງເອງ.
ຮັກສາການເຮັດວຽກຂອງທ່ານໄວ້
ບັນຊີຟຣີ ເພີ່ມຂໍ້ມູນໃສ່ບົດຮຽນທຸກບົດ, ບັນທຶກຂອງສິ່ງທີ່ທ່ານໄດ້ເຮັດແລ້ວ, ບັນຫາທີ່ທ່ານໄດ້ແກ້ໄຂໄວ້ໃນບ່ອນດຽວ ແລະ ຄູສອນທີ່ທ່ານສາມາດຖາມກ່ຽວກັບໜ້ານີ້. ຄະນິດສາດເອງແມ່ນເປີດໃຫ້ທຸກຄົນ, ບໍ່ວ່າຈະໄດ້ເຂົ້າສູ່ລະບົບ ຫຼື ບໍ່.
ລົງທະບຽນ
ចូល
ຕົວສະແດງທີ່ໃຊ້ຢູ່ທີ່ນີ້
ກົດຕົວອັກສອນໃດໜຶ່ງເພື່ອເບິ່ງຄວາມໝາຍເຕັມ, ຮູບ ແລະ ຕົວອັກສອນແຕ່ລະຕົວໃນມັນໝາຍຄວາມວ່າແນວໃດ.
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.
ຕົວສະແດງໃນໜ້ານີ້
ກົດເພື່ອເບິ່ງຄວາມໝາຍຂອງມັນ
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 ›
ຄໍາຖາມທີ່ຄົນຖາມ
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.
ບາງສ່ວນຂອງ ໜ້າ ນີ້ ຖືກປັບແຕ່ງຈາກ
Judson, Abstract Algebra: Theory and Applications (GFDL 1.3).
ຖືກປະສົມປະສານແລະອະທິບາຍຄືນໃໝ່ທີ່ນີ້; ຄວາມຜິດພາດແມ່ນຂອງເຮົາ.
ເພີ່ມເຕີມໃນ 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