External and Internal Direct Products -- Abstract Algebra 13

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[greencheesewheel]

Haven't commented in this channel before. Just wanted to pass by and tell you that your efforts don't go unnoticed. I'm a software dev from Argentina that always wanted to live from mathematics but never could. Couldn't even go to uni. So i've had to study on my free time from the very basics of algebra and real analysis, to group theory and topology. Online material is many times unhelpful because it assumes you've gone through the formal training college gives you. And even though i'm able to prove, for example, Cauchy's theorem for Abelian groups on my own I still have gaps in my knowledge that this videos help me notice. So, long story short, your videos have been an amazing help! Thank you, from the bottom of my heart! I may never get a degree in mathematics, but these type of channel means I can get closer to it than I imagined.

[GlenMacDonald]

My favorite part is "[40:57] when he casually says "...because 3 times 5 is 8 which is seven mod 8..." 😂
(Just kidding, Michael is great; I love these videos!)

[schweinmachtbree]

The fact that _m_ and _n_ are coprime is not needed at 26:50, so even if _m_ and _n_ aren't coprime we still get a homomorphism from U_ _mn_ to U_ _m_ × U_ _n_ ; it just won't be injective or surjective. If you package well-definedness and injectivity together then you have
" _a_ mod _mn_ = _b_ mod _mn_ ⇔ _mn_ | _a_ - _b_ ⇔ _m_ | _a_ - _b_ and _n_ | _a_ - _b_ ⇔ _a_ ≡ _b_ (mod _m_ ) and _a_ ≡ _b_ (mod _n_ ) ⇔ ( _a_ mod _m_, _a_ mod _n_ ) = ( _b_ mod _m_, _b_ mod _n_ )",
where the left direction of the second equivalence holds because _a_ - _b_ being a multiple of _m_ and _n_ implies it is a multiple of lcm( _m_, _n_ ) = _mn_ / gcd( _m_, _n_ ) = _mn_, and therefore we can use the coprimality condition neatly as
"phi( _a_ mod _mn_ ) = phi( _b_ mod _mn_ ) ⇔ ( _a_ mod _m_, _a_ mod _n_ ) = ( _b_ mod _m_, _b_ mod _n_ ) ⇔ _a_ ≡ _b_ (mod _m_ ) and _a_ ≡ _b_ (mod _n_ ) ⇔ _m_ | _a_ - _b_ and _n_ | _a_ - _b_ ⇔ *_mn_** = lcm( **_m_**, **_n_** )* | _a_ - _b_ ⇔ _a_ mod _mn_ = _b_ mod _mn_ "

[Hipeter]

Is the lcm inequality at 20:38 valid? For example, 3<4 but lcm(2, 3)=6>4=lcm(2, 4).

[finnr]

In the last step in the proof that the homomorphism U_mn -> U_n × U_m is injective, one could simply argue with the Chinese Remainder Theorem.

[jordimartinez]

How do you prove that talking non-stop for 40 minutes is exhausting? 40:57🤯

[M.athematech]

At 27:40 your presentation of the well-definedness argument makes the mistake of already using the definition just before having reached the conclusion that is well defined.