Cosets and Lagrange’s Theorem - The Size of Subgroups (Abstract Algebra)

to fully understand a group G you need to take it apart and study the pieces these pieces are called subgroups there are smaller groups contained in G but where do you begin to look for subgroups how do you find them if there are any by using objects called Co sets we can create a simple rule that dramatically narrows down the possible list of subgroups this result is called the garages theorem today we will prove this theorem and show how useful it can be quick note for the rest of this video we will only be talking about finite groups a subgroup of a group G is a subset that's also a group if H is a subgroup of G we write it like this every group has at least two subgroups itself and the trivial group the trivial group consists of a single element the identity element this is similar to how every positive integer has at least two designers itself and one but these subgroups are fairly dull we want to see if G has a subgroup other than these two something more interesting the map location Joseph Leela grunge found a useful test which we now call Lagrange's theorem he discovered that if H is a subgroup of a finite group G then the order of H divides the order of G recall that the order of a group is a number of elements in the group and it's denoted by the absolute value symbol with this notation we can write Lagrange's fear and more compactly this is a powerful result it says that subgroups can't be any old sides there are strong restrictions on the possible subgroups of G before proving this theorem let's see just how useful this is imagine you have a group G with 323 elements you can factor this at 17 times 19 this means the only positive integer is dividing 323 r1 1719 and 323 so any subgroup of G must have order 1 1719 or 323 earlier we mentioned that every group has at least two subgroups itself and the trivial group the group itself has order 323 and the trivial group has ordered what this means if G has any other subgroups their orders are 17 or 19 the list of suspects has shortened dramatically this is a good moment to pause and not get overly excited about Lagrange's theorem we need to learn its limitations 17 and 19 divide the order of G does lagravis theorem say that there are subgroups of these orders no Lagrange's theorem is an if/then theorem which means the converse may not be true just because the number divides the order of G there's no guarantee that a subgroup of this size will exist in this case it just so happens that G does have subgroups of order 17 and 19 but there are groups where this isn't the case for example there's a group of 12 elements called a for the alternating group on 4 elements the divisors of 12 are 1 2 3 4 6 and 12 so any subgroup must have one of these orders however even though the number 6 divides 12 this group does not have a subgroup of order 6 but it does have subgroups of the other possible orders this is a cautionary tale on the limitation of Lagrange's theorem let's now roll up our sleeves and get to work proving this theorem suppose we have a random finite group G with order n to start let's check the two easy cases of the trivial group and the group itself the trivial subgroup has order 1 and 1 divides in next the group itself has order N and n divides n sola gracia theorem works for both of these cases now suppose that G has a proper subgroup H that's not the trivial group recall that a proper subgroup is just a subgroup that's smaller than G 1 that's a proper subset will represent a group G by a rectangle an H as a small square inside G because H is a group it contains the identity element so the identity element for both G and H lies inside the H square here is a construction take an element G one that lies outside of H next let's look at the set G 1 times H this is the set you get when you multiply every element in H by G 1 on the left remember we never said that G is abelian so the order matters this set is called a left coset because we are multiplying the elements of H by G 1 on the left the key observation is that H n t18 have no element in common their intersection is the empty set here's why let's assume that there is an element in both H and G 1 H this means G 1 times H I would equal H J for some elements H I and H J in the subgroup H if you multiply this equation on the right by the inverse of H I then simplify you get G 1 equals H J times H I inverse since H is a subgroup the right hand side is an H so G 1 is in H but this is a contradiction we specifically chose u 1 to not be in H so our assumption must be wrong and the two sets do not overlap we're going to repeat this process until we get to the point there's no element in G that's not in a coset pick an element G 2 that lies outside both H and G 1 H again we'll look at the left coset g 2h using the same argument as before H and G to H do not overlap next we'll show that G 1 H and G to H do not overlap either assume they do overlap then G 1 H I equals G 2 H J for some elements H I and H J in the subgroup H if you multiply both on the right by H J inverse simplify you get G 1 H I H J inverse equals G 2 but H I times H J inverse is 1/8 called it H K which means G 1 H K is in the left coset to UNH this means that G 2 is in the coset 2 u 1 H but we specifically chose G 2 to be outside v18 so our assumption was wrong and they do not overlap continue this process until you get to the point that there is no element look that's not in a closet the result is we have partition G into a collection of non-overlapping left cosets there is one more thing all the co sets are the same size there is a short proof for this consider a left coset GH suppose the set contains a duplicate element this means G h1 equals G h2 for different elements h1 and h2 but if you multiply on the left by G inverse you see that h1 equals 82 this means that the size of each left coset is equal to the order of H if the order of H is D then each co set contains the elements we've now managed to split G into non-overlapping left cosets a word of caution H is a subgroup while the other closets are not groups they're just set this is because only H contains the identity element and to be a group you have to have an identity element let's call the number of Co sets K the number of Co sets is called the index of H in G and is written like this so we have B times K equals and the order of G in other words D divides n this means the order of H divides the order of G this proof Lagrange's theorem one last thing throughout the proof we worked with left cosets but you can repeat the proof using right closest to for abelian groups the left cosets and white coasts are the same but for non abelian groups this may not be the case either way the proof is this pick a subgroup keep making Co sets until the group G is covered once you show that Co sets don't overlap and are all the same size you've proven Lagrange's theorem socratic of partitions our videos into non-overlapping sets called 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Channel: Socratica

Views: 405,547

Rating: 4.9619555 out of 5

Keywords: Socratica, ​ ​SocraticaMath, ​ math, maths, mathematics, ​abstract​ ​algebra, group, groups, group theory, coset, cosets, lagrange’s theorem, lagrange, subgroup, subgroups, educational video, modern algebra, higher math, advanced math, college, university, STEM

Id: TCcSZEL_3CQ

Length: 9min 18sec (558 seconds)

Published: Mon Mar 20 2017

Reddit Comments

[u/boyobo]

I liked the presentation and the animations. I had a look at some of their other videos, it seems to me that the slides and the voice are already very effective on their own, in my opinion having the camera on the gesticulating/emoting actress bit distracting. Then again maybe it helps other people stay engaged.

[u/strategyguru]

I thought this was explained pretty nicely with rigor and the video deserves a few more views...

[u/RiemannCalculator]

Gotta say, this was very well made. I'm pretty rusty with my math but the way she explained the proof (alongside those smooth animations) made the concept easy to understand.

[u/failedgamor]

Just in time as I prepare for my exams!