Abstract Algebra, Lec 6B: Subgroup Tests, Cyclic Subgroups, Center of a Group, Cyclic Groups

(0:00) Cayley table for Z6 rearranged to emphasize that {0,2,4} is a subgroup of Z6 under addition mod 6.
(3:53) Cyclic subgroup of G generated by an element "a".
(6:15) Analogy with vector spaces: one-dimensional subspace (spanned by one vector) in the vector space.
(7:45) Use the one-step subgroup test to prove that the cyclic subgroup generated by "a" is truly a subgroup (need to test that the given set is nonempty first).
(14:36) Two-step subgroup test.
(15:26) Finite subgroup test.
(15:55) Infinite groups can have finite subgroups (even beyond the trivial subgroup).
(17:18) {1,i,-1,-i} is a cyclic subgroup of order 4 of C - {0}: it has i = sqrt(-1) as a generator.
(18:50) Center of G is a subgroup of G (the center is the set of all elements of G that commute with everything in G) and idea of how to use the two-step subgroup test to prove it.
(27:32) Definition of a cyclic group and consider examples: Z is cyclic, Z6 is cyclic, Zn is cyclic (1 and n-1 are always generators, though there could be other generators, especially when n is prime), U(10) = {1,3,7,9} is cyclic (under multiplication mod 10), but U(8) is not cyclic (under multiplication mod 8).
(31:26) Note that cyclic groups must be Abelian, so non-Abelian groups are not cyclic.

AMAZON ASSOCIATE
As an Amazon Associate I earn from qualifying purchases.

[biskoot_pc]

So well explained!
IDK why this video has so few views

[youngkim]

VERY great lectures Professor. Also I have a question:

Is there an error @ 31:05 when you try to disprove that u(8) is cyclic? i'm not sure how showing that 3^2, 5^2 and and 7^2 = 1 under multiplication mod 8 disproves "cyclicness" if thats a word haha.

[harshvardhansingh]

Sir, just to clarify some doubts at 13:42, we have taken a subset of G namely H which is such that each element in H, when raised to the powers equal to all the integers, the set thus obtained is itself a group and hence a sub-group of G .
Essentially, a sub-group by definition is a group formed from the elements of a larger group under the same binary operation as the group.
My doubt is this :- Can i take an element from a Group G say 'a' and perform operation on it as X={a^n | n ∈ Z}and be sure that the elements in set X will always be present in the group G ? Will such a set X formed be a subgroup of G ?

[harshvardhansingh]

@Bill Kinney Sir, in the subgroups Cyclic subgroups and Centre of a group which forms a subgroup with respect to group G, the binary operation
is multiplication.I am asking this because this is not explicitly mentioned and i have seen couple of examples here and there where to prove that a group is a subgroup of a larger group, generally the examples proceed with binary operation being multiplication although it is not mentioned anywhere.
Thanks !!