Visual Group Theory, Lecture 1.4: Group presentations

[@ykge8877]

For the last group at 18', using associativity, ba=a^2b=(aa)b=a(ab)=a(b^2a)=abba=(ab)(ba), so ab=e. So a=b^-1. Hence e=ab=b^2a=(bb)a=b(ba)=b. And a=e as well.

[@hellshulk]

These videos are very very helpful. Thanks for putting up these.

[@rage4dorder]

This is great! Looking forward to your upcoming book!

[@valor36az]

Excellent set of lectures

[@jongraham7362]

At 9:47, I think you meant to say fr^2 = rf... and then you can get the other one from that.

[@nathanielvirgo]

8:50 I don't get it. How do we know that these relations uniquely define this group? I.e. how do we know there isn't some other group that also has these relations, but also some other ones?

[@4sety]

Wouldn't that just make the group in question a subgroup of whatever group happens to entail other relations/generators?

[@henlofrens]

is abelianality then always denoted in the relations part of the group representation?

[@edvogel56]

Thank you for these well illustrated and paced lectures.

[@wesleysuen4140]

9:25 I bet you wanted to write rf=fr^2 instead.

[@manhtuannguyen9685]

Hello Professor Macauley, Isn't it r is 120 degree clockwise rotation? Could you look at 8.26, at the outer circle of flipped object? Because it appears to me that the flipped object is rotated counter-clockwise. Could you explain why or point out where I have been wrong? You made good lectures. Thank you.

[@superadey7215]

Thank you for this!

[@gleedads]

There were two occasions in this lecture where you let a = e and b b = e. But doesn't this violate one of the central tenets of set theory that every element of the set must be distinct?

[@marcoantonio7648]

Haha, I knew it!