Abstract Algebra | The dihedral group

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Күн бұрын

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@franciscodanielquiroz9904 3 жыл бұрын

Why nobody is talking about how cool it is that he snaps his fingers to clean the board? As mathematics teacher I create my own videos as well and this gives me great ideas! Love this video.

@cycklist 5 жыл бұрын

Your channel is so good. It's wonderful to watch this more advanced stuff; it takes me back to my undergraduate days and all those happy memories. Best wishes to you from the UK.

@paul21353 3 жыл бұрын

This absolutely is the same for me. These video's take me back to my first years studying math at my uni and also it brings the joy of understanding the material better than back in those early years.

@xoppa09 24 күн бұрын

9:48 We can also show r ∘ s = s ∘ r^(n-1) by algebraically computing (r ∘ s)^ -1 and then use the fact that r ∘ s is a reflection (a product of a reflection and a rotation is a reflection) and the fact that every reflection is its own inverse. You can also test this statement on a small regular pentagon or regular hexagon. Claim : r ∘ s = s ∘ r^(n-1) Proof: First we will show (r∘ s) ^-1 = s ∘ r^(n-1) . (r ∘ s)^-1 = s^-1∘ r^-1, by shoe socks theorem = s ∘ r^-1, replacing s^-1 with s (since s is a reflection s^-1 = s) = s ∘ r^(n-1) , since r^-1 = r^(n-1). Because (r∘ s) ^-1 = r ∘ s since r∘ s is a reflection , we can substitute r∘ s for (r∘ s)^-1 and we get the result r ∘ s = s ∘ r^(n-1).

@basakatik4770 4 жыл бұрын

Finally I got the concept totally! Thank you very much for this clear and wonderful explanation!

@paul21353 3 жыл бұрын

The drawings starting around 7:50, combining rotation and reflexion of an n-gon presuppose that n is odd. Strictly speaking you should check that the result is the same when n is even.

@xoppa09 24 күн бұрын

5:11 I don't think he meant to say s and r commute since s ∘ r = vertical flip and r ∘ s = horizontal flip. Also the dihedral group D_4 (some books denote it as D_2*4) is a nonabelian group. I think he meant to say, how can we get r ∘ s to look like s ∘ r^k where k is some integer , which is the form used in the representation of the group i.e. he is trying to find the relationship r ∘ s = s ∘ (some rotation). And to do that we can use the formula r ∘ s = s ∘ r^(n-1), which he explains how to get geometrically. Alternatively you can use the geometric fact r ∘ s ∘ r = s, because when you rotate, reflect , and rotate again , that last rotation is a mirror image rotation (a rotation in mirror world) - so it undoes the first rotation and you are back to s, your original reflection. And then you can multiply both sides by r^(n-1) to get the result (and the fact that r^n = e) .

@xoppa09 24 күн бұрын

The pentagon and the hexagon are two good case examples to generalize or infer how we actually go about finding the n-reflections of an n-gon, for n odd and n even. Rotation is much easier.

@abnereliberganzahernandez6337 Жыл бұрын

This Is my Man right there! One of my favorite videos all Time.

@liranekm 3 жыл бұрын

OMG This is gold . Love my math instructor but me not taking number theory has been a set back. THANK YOU

@xoppa09 20 күн бұрын

15:35 what does that symbol v underlined mean? base case?

@xoppa09 20 күн бұрын

There is a typo at 16:58, he literally said the correct thing and wrote the wrong thing , which is interesting. All good, great teacher. Also note that n is some positive integer (e.g. n = 6) and is fixed in the proof, so we are doing induction on the k between 1 and n-1 inclusive.

@余淼-e8b 3 жыл бұрын

The previous video is kzbin.info/www/bejne/qJLTi51vrtamhNk

@ikramedaqaq861 Жыл бұрын

thank you

@SanketAlekar 2 жыл бұрын

At 10:00, it should be n-1 clockwise rotations (r^n-1) followed by a reflection that fixes 1 (s) to be consistent. What you did was n-1 counter-clockwise rotations (r^(n-1))^(n-1), following by a reflection that fixes n (which is not s).

@lancelofjohn6995 3 жыл бұрын

nice,in the end the r^(k+1)=s*r^n*r^n-(k+1)

@eduardohenriquerodriguesdo6103 4 жыл бұрын

another proof of rs=sr^(n-1): note that they are inverses. Because they are both reflections,it must be the case that they are equal.

@MichaelPennMath 4 жыл бұрын

Nice and quick!

@余淼-e8b 3 жыл бұрын

Brilliant!

@georgettebeulah4427 5 жыл бұрын

I love this explanation I can relate with it a lot thank you for loading on time

@MichaelPennMath 5 жыл бұрын

Thanks

@余淼-e8b 3 жыл бұрын

Love your channel so much. Thanks for sharing.

@АннаКрылова-х1п 12 күн бұрын

Thanks ❤

@joetursi9573 3 жыл бұрын

We must be careful not to confuse rotation as being restricted to it's common definition of rotating through 2pi or 360 degrees. In this context a rotation means a "motion." If not then writing that the number of rotations=2(pi) K/n where k is )=k-< or equal to n-1 is confusing. Example: Set n=3 (a triangle) we have that 2(pi) k/3- the number of rotations, implying K=9/2(pi) which is about 3/2( not even an integer in the set[0,N-1} . It's about one and a half rotations which certainly note equal to what is correct:3 .