👍Thanks, and a temptation - For every odd prime p, any Ring with 2p elements is Commutative (∵ If the Addition Group is Cyclic, the Multiplication Commutes, and There's but one Abelian, viz. the Cyclic Group of this order) Any independent Corroboration / Refutation of this tempting claim, will be appreciated😃
Hey Doctor Bob. I found your channel quite useful! Thank you a lot! Could you post a lecture about Wreath products? I would be quite grateful for that!! Thanks in advance!!
Could you expand a bit on what you did on 9'42'', erase the parentheses and the comma? The direct product of two sets (groups) becomes the product of two subgroups.
As defined, D_{2n} is not a product of groups. We can find an H and N so that H x N (semi direct) is isomorphic to D_{2n}. Dropping the comma here is just the isomorphism (c^i, r^j) -> c^ir^j.
@@fsaldan1 That would be true for a direct product (for all elements). For semi direct product, you need to account for conjugating by c. You can read the semi direct product rule off the previous board. The map promises G=HN as a set, but you can't read multiplication off that. (Maybe before I mean 'bijection' instead of 'isomorphism')
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this is the most useful channel on youtube!
these videos are keeping me from failing!!!
You're welcome! I'm taking a break from algebra once I finish the Field Theory playlist, so it wouldn't happen for a long while.
I wish! Thanks for the kind words.
Great Video. Helped me out a lot!
👍Thanks, and a temptation -
For every odd prime p, any Ring with 2p elements is Commutative
(∵ If the Addition Group is Cyclic, the Multiplication Commutes, and There's but one Abelian, viz. the Cyclic Group of this order)
Any independent Corroboration / Refutation of this tempting claim, will be appreciated😃
Hey Doctor Bob. I found your channel quite useful! Thank you a lot! Could you post a lecture about Wreath products? I would be quite grateful for that!! Thanks in advance!!
First of all, thank you very much for this lecture video. But can you show us in the matrix form. Thanks in advanced.
Could you expand a bit on what you did on 9'42'', erase the parentheses and the comma? The direct product of two sets (groups) becomes the product of two subgroups.
As defined, D_{2n} is not a product of groups. We can find an H and N so that H x N (semi direct) is isomorphic to D_{2n}. Dropping the comma here is just the isomorphism (c^i, r^j) -> c^ir^j.
@@MathDoctorBob Ok, thanks. But for this to be an isomorphism one needs (c^ir^j)(c^kr^l) = (c^(i+k)r^(j+l)). Is that true?
@@fsaldan1 That would be true for a direct product (for all elements). For semi direct product, you need to account for conjugating by c. You can read the semi direct product rule off the previous board. The map promises G=HN as a set, but you can't read multiplication off that. (Maybe before I mean 'bijection' instead of 'isomorphism')
@integralmath As a group theorist, well, I hear that one a lot. :) - Bob
Hi Dr. Bob. Why πy(x)=yxy^-1? Why does π(y) have to be the conjugation, instead of other automorphisms?
Need some help - is this in the video, because I'm pretty clear about using general automorphisms?
@MathDoctorBob that's *almost* hard to imagine.
4:05 : By pi=I, you mean the constant homomorphism mapping every h to the identity on N?
Great videos by the way. Thanks for making these!
Your welcome! I guess I'm back! :)
It's been eight years: pi=I says for any x, pi(x)n=n. So yes!
@paul1964uk Thanks! I annotated a correction. Math never sleeps. - Bob
Is there a typo at 6:24? Should the product not be "(e sub H, e sub N)" rather than "(e sub H, e sub G)"?
Thank you!!.
If the two grapes are abelian, then . . .
Sorry, I couldn't help myself!
Frod h ye sb log😈