Abstract Algebra Chapter I
Name: Ans-Wain
Totally 110 points, including 10 bonus ones. Please nish in 50 minutes.
1. (20 points) Consider the following binaryr algebraic structures.
(A) Determine which ones are groups, which ones are not.
(B) If a. struct
Abstract Algebra Chapter II
Name: A HS W 8!"
Totally 110 points, including 10 bonus ones. Please nish in 50 minutes. Here
(3) Sn denotes the symmetric group of n letters;
(b) The order of a finite group is the number of elements in thie group:
(c) The ord
1.6. CYCLIC SUBGROUPS
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1.6
Cyclic Subgroups
Recall: cyclic subgroup, cyclic group, generator. Def 1.68. Let G be a group and a G. If the cyclic subgroup a is finite, then the order of a is | a |. Otherwise, a is of infinite order.
1.6.1
Elementary Prope
2.3. COSETS AND THE THEOREM OF LAGRANGE
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2.3
Cosets and the Theorem of Lagrange
We always assume that H is a subgroup of the group G.
2.3.1
Cosets
Def 2.33. Let H be a subgroup of G. Given a G, the subset aH = cfw_ah | h H of G is the left coset of H co
3.4. (III-16) GROUP ACTION ON A SET
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3.4
3.4.1
(III-16) Group Action on a Set
Group Action
We have seen many examples of group acting on a set. Ex 3.54. The group D4 of symmetries of a square. Ex 3.55. The symmetric group Sn and the alternating group An
Chapter 4
7. Advanced Group Theory
It is important to build up the correct visions about things in a group, a homomorphism, or so.
4.1
VII-34. Isomorphism Theory
Thm 4.1 (First Isomorphism Theorem). Let : G G be a group homomorphism with kernel K . Let K
4.2. VII-35. SERIES OF GROUPS
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4.2
VII-35. Series of Groups
To give insights into the structure of a group G, we study a series of embedding subgroups of G.
4.2.1
Subnormal and Normal Series
Def 4.10. A subnormal (or subinvariant) series of a group G is
4.3. 36. SYLOW THEOREMS AND APPLICATIONS
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4.3
36. Sylow Theorems and Applications
The structures of finite abelian groups are well classified. The structures of finite nonabelian groups are much more complicate (Think about Sn , An , Dn , etc). Sylow th
01.08 (1 + i)3 = 1 + 3i + 3i2 + i3 = 1 + 3i - 3 - i = -2 + 2i 01.17 z 4 = -1 = ei . So z = ei(/4+2k/4) for k = 0, 1, 2, 3. Note that ei = cos + i sin 01.38 eia eib = ei(a+b) . By Euler's formula, (cos a + i sin a)(cos b + i sin b) = cos(a + b) + i sin(a +