Abstract Algebra Exam 2 Review Problems and Solutions

#abstractalgebra #abstractalgebrareview #grouptheory

Links and resources
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(0:00) This is about intermediate group theory
(0:32) Normal subgroup definition
(2:44) Normal subgroup test
(4:09) Lagrange's Theorem
(7:06) Apply Lagrange's Theorem: find possible orders of subgroups of a group of order 42
(9:11) Are U(10) and U(12) isomorphic or not?
(13:52) Number of elements of order 4 in Z2 x Z4 (external direct product of Z2 and Z4)
(18:11) Number of elements in HK, where H and K are subgroups of G (if H and K are normal subgroups of K, then HK = KH and HK will be a subgroup of G, called the join of H and K)
(20:24) Factor group coset multiplication is well defined (Quotient group coset multiplication is well defined). Where is normality used?
(26:59) Cauchy's Theorem application: If G has order 147, does it have an element of order 7 (if p is a prime that divides the order of a finite group G, then G will have an element of order p).
(28:45) Groups of order 2p, where p is a prime greater than 2
(29:47) Groups of order p, where p is prime
(30:21) G/Z Theorem
(31:50) The functor Aut is a group isomorphism invariant (if two groups are isomorphic, their automorphism groups are isomorphic)
(32:49) Is Aut(Z8) a cyclic group?
(33:57) Is Z2 x Z5 a cyclic group? How about Z8 x Z14?
(34:36) Order of R60*Z(D6) in the factor group D6/Z(D6)
(37:19) Abelian groups of order 27 and number of elements of order 3
(42:28) Prove: If a group G of order 21 has only one subgroup of order 3 and one subgroup of order 7, then G is cyclic.
(47:28) A4 has no subgroup of order 6 (the converse of Lagrange's Theorem is false: the alternating group A4 of even permutations of {1,2,3,4} has order 4!/2 = 12 and 6 divides 12, but A4 has no subgroup of order 6)
(55:32) Elements and cyclic subgroups of order 6 in S6 (S6 is the symmetric group of all permutations of {1,2,3,4,5,6} and has order 6! = 720)
(1:02:18) U(64) isomorphism class and number of elements
(1:04:58) Number of elements of order 16 in U(64)
(1:07:11) Order of 3H in factor group U(64)/H, where H = (7) (the cyclic subgroup of U(64) generated by 7)
(1:09:53) Preimage of 7 under a homomorphism φ from U(15) to itself with a given kernel (ker(φ) = {1,4} and given that φ(7) = 7)
(1:12:15) Prove the First Isomorphism Theorem (idea of proof)

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[scoobydoofan]

Hi Bill, thank you for making this very helpful video. For question 17 I think there are actually 8 elements of order 3 in Z_3 x Z_9, not 6. in the case that |a| = |b| = 3, there are 4 elements that satisfy this requirement

[eddiejennings]

Thank you again, professor. I concur with your correction on question 17; You provide extremely thorough explanations. I also really like it when you draw connections and highlight differences in ideas. Nice proof/quick answer problems. All the best to you and yours.

[noahsammond]

Thank you for this. My graduate abstract algebra final is tomorrow and I think I understand things much better now.

[cioewsl]

Thank you. This was really helpfull. If you have time, please do more videos like this.