Cyclic Groups -- Abstract Algebra Examples 7

[@knowledgedose1956]

Thank you for these videos. One of the most well explained course with detailed explanations

[@Happy_Abe]

@7:40 aren’t all of them generators for

[@schweinmachtbree1013]

All of them except e = r^0, yeah

[@Happy_Abe]

@@schweinmachtbree1013 awesome thanks

[@Happy_Abe]

@@schweinmachtbree1013 so he made a mistake there I guess

[@schweinmachtbree1013]

@@Happy_Abe yes there are mistakes in most of these example videos

[@Happy_Abe]

@@schweinmachtbree1013 thanks for the answers

[@wenzhang365]

These set of excercises is quite helpful, thank you.

[@donaldfalkenhagen7408]

At 5:18, another way to find all generators of

[@psolien]

I would say it's way easier to state ord(r)=21 and treat it as a subroup of 😉

[@JasonArezki]

Awesome stuff, do you give one to one lessons ?

[@jacoberu-q2w]

at 5:40, why reduce mod 21 in D_21? shouldn't it be 42 since D_n is order 2n? also at 7:30, some arrows are pointing the opposite direction. at 14:10, why only the small prime factors of 124? < < <

[@luminatihd2895]

Good Video, but I think could prove the last claim in a different way (not that I dislike your proof): Suppose q ∈ Q, q = a/b, a,b ∈ Z and gcd(a, b) = 1, a>b => 1

[@schweinmachtbree1013]

An improvement on your proof: If ℚˣ were cyclic, generated by q, then: - q has to be negative since if q were non-negative then all its powers would be non-negative, not reaching all of ℚ. also clearly q ≠ -1. - but then -q is not in : if q^n = -q then q^(n-1) = -1 so n-1 would have to be odd, but then q^(n-1) = -1 ⇒ q = -1, contrary to the above.

[@luminatihd2895]

That's a good one. I completetly neglected the negative numbers in my proof, so yours is propably also more complete.

[@schweinmachtbree1013]

@@luminatihd2895 Your proof suffices actually, using the fact that any subgroup of a cyclic group is cyclic: if ℚˣ were cyclic then ℚ₊ = {x ϵ ℚ | x > 0} would be too, so ℚ₊ is not cyclic ⇒ ℚˣ is not cyclic :)