302.1B: Dihedral and Symmetric Groups

Review of the elements and operations of dihedral and symmetric groups.

[lordbossharrow]

you're way better than my lecturer. concise and clear, thank you! :)

[ahmadharvey]

This is a great lecture! However, shouldn't there only be 6 4-cycles? Are not (2134) and (1342) the same cycle?

[DaeGlyth]

Would it be accurate to visualize t as a "flip."  For example, if the square were a cut-out piece of paper on a table and I flipped it over (horizontally) and placed it back face down.  In fact, I think just a piece of paper with each corner labeled (1, 2, 3, 4) would really help to "see" all of these in action.  

Trying to explain what I've learned so far from these two videos... please correct if this is inaccurate:

So if the group is D4 is the finite set {r, r^2, r^3, e, t, tr, tr^2, tr^3} and * is the binary operation on D4, then the composition r * r^2 would equal r^3 (which is already in the set, hence closure). r * e = r, t * r = tr, etc.

[shannaflores]

For the symmetric group, you counted the identity permutation twice. You first counted it as a 4 cycle (1234), then said to add the identity to the number of elements. It has 1+6+8+7+3 = 25 but should be just 6+8+7+3 = 24.

[MatthewSalomone]

Note the following typo: since (2143) = (1432), there are only 6 4-cycles in S4. This explains the phony arithmetic.

[anishaq]

WOW!! Love your videos! I was going through Dummit, Saracino for understanding these. Your explanation is really great and I have clarified what I couldn't reading those books. Thanks. Almighty bless you!

[ali]

Excellent lectures. Thank you
Question. 8:50----9:38. The elements of S(six), as you have written them, I counted 25 elements rather than 24. Do you not count the Identity?

[bonbonpony]

04:39 It's quite easy to see that some operation is non-commutative. The more interesting question, hovewer, is WHY is it non-commutative ;> what *mechanism* makes it non-commutative. And how to find it out without checking all possible products until some of them will break the symmetry :q (because it might be the case that _some_ products are commutative, while just a bunch of them isn't, breaking the commutativity of the entire operation).

[sanjursan]

Maybe you should have labeled the corners from the very beginning. I was led to believe that the symmetry of a rotation by ninety degrees left the square indistinguishable from its starting position. (1:20 - 1:35) But if the corners had been labeled, then this would not qualify as a symmetry. Or did I just confuse myself again? Lol. Anyone else?

[alojzybabel]

08:20 The remaining ones are actually products too :> For example, (124) is actually (124)(3), that is, a product of the three-cycle (124) with a 1-cycle (3) which consists only of 1 element that stays at its position unchanged ;)

[aliharkatti]

Hello thank you for this video but 1+6+8+7+3=25 not 24

[PunmasterSTP]

Dihedral and symmetric? More like "Dang cool? You bet!" 👍

[DaeGlyth]

Regarding, "Symmetric Groups."  If these are really permutations... then why are they called symmetric at all and not "Permutation Groups"?  I'm not sure how they are "symmetric" at all per se.  It wouldn't be so bad if not for the confusion of symmetric being used in a very different context to explain dihedral groups.

I'll give a pass on why commutative groups are called Abelian, since they are named after Niels Henrik Abel.