Maybe you can address the fact that for all n, there are more groups of order 2^n than groups of order k for all k < 2^n. And that there are A LOT more groups of order 2^n than most other orders, which has always fascinated me.
What's going on here is that the number of groups of order k is not so much determined by the size of k but moreso by the _number of prime factors_ of k, so the relation "
@@schweinmachtbree1013 I don't think it's right to say < is "not the right relation" when the statement I said is true. However, I do agree that generally when looking at arithmetic objects, the "divides" relation is good to consider as well. And indeed, there is a lot going into it, which is why I think it'd make a good video! It's a good overarching goal to introduce a lot of important group theory - from cyclic groups, to direct products and semidirect products, to FTFGAG, and maybe even the 26 sporadic groups (with their wonderful history!). I saw Robert Griess talk about his work with the Monster Group and I think it'd be a wonderful tale for Michael to go over.
Using some known results can simplify some cases. For example in the case 2, after taking the 3 elements x, y and z that Penn considered. You can prove that G is the direct product of , and and therefore you get that G is isomorphic to (Z_2)^3. To prove this it suffices to show that these 3 subgroups are normal in G (but you know this since G is abelian) and to show that the intersection of one of them with the subgroup generated by the other 2 is {e} (but you know this since they are 3 different non identical elements and since xy is not z). So, since =G, you get that G is the direct product of these 3 subgroups.
another way i guess is to just trivially inspect: xy, xz, and yz are taken to all be distinct from x, y, z, e by construction, and so you can check that xyz must multiply distinctly to all the previous 7 mentioned elements, which makes 8 elements
I like the representation theory of groups of order 8 as well. If I remember correctly its the lowest order where you have non-isomorphic groups with the same character table. It’s a nice example that representations don’t completely eliminate the need for these types of group theoretic analyses.
Thank you for posting this. I was self studying Abstract Algebra and I got about 20% of the way through but I was really struggling with rigorously structuring some of the proofs. Hopefully this will inspire me to get back on track.
As with most things, what's most important is not giving up and practice practice practice. as you do more you'll start to realise that what used to be difficult is now second nature. If you don't already know these proofs I would suggest them because I find them quite accessible and together they give a good intuition for functions: - the composition of two injective functions is injective - the composition of two surjective functions is surjective - therefore the composition of two bijective functions is bijective - a function is bijective if and only if it has an inverse function - function composition is associative - the identity function on a set is a bijection - the inverse function of a bijection is a bijection - from the previous three results, the collection of bijections from a set to itself forms a group (when the set is finite, say of size n, this group is isomorphic to the symmetric group S_n). Michael also has a playlist on proof writing on his main channel if you'd like to check that out.
What's the intuition for landing on Z_2 x Z_2 x Z_2 as a candidate for an isomorphic group? Just wondering what intuitively makes that pop out (especially for an undergrad through 1/3rd of an abstract algebra class) Thanks :)
Because Z_2 x Z_2 x Z_2 is abelian and every non-identity element has order 2, which is easy to see because 0+0 = 0 and 1+1 = 0 in Z_2 so if you add any element (i, j, k) to itself you get (i+i, j+j, k+k) = (0, 0, 0). If you like geometrical visualisations then Z_2 x Z_2 x Z_2 is the symmetries of a cuboid whose three sidelengths are all different. This generalises the Klein Four group Z_2 x Z_2 which is the symmetries of a non-square rectangle - Z_2 x Z_2 is generated by the two reflections of the rectangle and Z_2 x Z_2 x Z_2 is generated by the three reflections of the cuboid.
6:57 It's actually because they have the same size _and_ they are finite sets. 8:59 This is the standard method of showing a _homomorphism_ is injective. Obviously an arbitrary function doesn't necessarily have a kernel defined (at least not a _group_ kernel; there is a more broad definition of kernel which applies to categories).
I think those are both nitpicking: - I usually only say "size" for the cardinality of a finite set, so maybe michael uses this convention too (there's no need to use big/fancy words when not necessary - same goes for "transpositions" which should just be called "swaps"), and anyway he says that the sets are finite in the very next sentence. - because kernels aren't defined for arbitrary functions, by saying "showing something is injective by ... it's kernel" it is implicit that the "something" has a kernel, so is a homomorphism and not just an arbitrary function
Maybe you can address the fact that for all n, there are more groups of order 2^n than groups of order k for all k < 2^n. And that there are A LOT more groups of order 2^n than most other orders, which has always fascinated me.
What's going on here is that the number of groups of order k is not so much determined by the size of k but moreso by the _number of prime factors_ of k, so the relation "
@@schweinmachtbree1013 I don't think it's right to say < is "not the right relation" when the statement I said is true. However, I do agree that generally when looking at arithmetic objects, the "divides" relation is good to consider as well.
And indeed, there is a lot going into it, which is why I think it'd make a good video! It's a good overarching goal to introduce a lot of important group theory - from cyclic groups, to direct products and semidirect products, to FTFGAG, and maybe even the 26 sporadic groups (with their wonderful history!). I saw Robert Griess talk about his work with the Monster Group and I think it'd be a wonderful tale for Michael to go over.
Using some known results can simplify some cases. For example in the case 2, after taking the 3 elements x, y and z that Penn considered. You can prove that G is the direct product of , and and therefore you get that G is isomorphic to (Z_2)^3. To prove this it suffices to show that these 3 subgroups are normal in G (but you know this since G is abelian) and to show that the intersection of one of them with the subgroup generated by the other 2 is {e} (but you know this since they are 3 different non identical elements and since xy is not z). So, since =G, you get that G is the direct product of these 3 subgroups.
another way i guess is to just trivially inspect: xy, xz, and yz are taken to all be distinct from x, y, z, e by construction, and so you can check that xyz must multiply distinctly to all the previous 7 mentioned elements, which makes 8 elements
I like the representation theory of groups of order 8 as well. If I remember correctly its the lowest order where you have non-isomorphic groups with the same character table. It’s a nice example that representations don’t completely eliminate the need for these types of group theoretic analyses.
Could you tell me more? Thank you
Thank you for posting this. I was self studying Abstract Algebra and I got about 20% of the way through but I was really struggling with rigorously structuring some of the proofs. Hopefully this will inspire me to get back on track.
As with most things, what's most important is not giving up and practice practice practice. as you do more you'll start to realise that what used to be difficult is now second nature. If you don't already know these proofs I would suggest them because I find them quite accessible and together they give a good intuition for functions:
- the composition of two injective functions is injective
- the composition of two surjective functions is surjective
- therefore the composition of two bijective functions is bijective
- a function is bijective if and only if it has an inverse function
- function composition is associative
- the identity function on a set is a bijection
- the inverse function of a bijection is a bijection
- from the previous three results, the collection of bijections from a set to itself forms a group (when the set is finite, say of size n, this group is isomorphic to the symmetric group S_n).
Michael also has a playlist on proof writing on his main channel if you'd like to check that out.
Speaking as a Physicist, I love examples! Thank you!😍
thanks so much
Takes me back. Still able to follow well, I loved group theory when I studied in the 80s. Maybe do order 12 soon?
For the proof that x^2=e for all x => abelian:
e=(xy)^2=xyxy => y=xyxy^2=xyx => yx=xyx^2=xy
feels a bit quicker
you can also use that x = x^{-1} for all x, so for all x and y one has yx = y^{-1}x^{-1} = (xy)^{-1} = xy.
Pure joy
the first three cases are just abelian cases, so we can use the Fundamental Theorem of Finite Abelian Groups.
Hey bozo he explicitly stated that they were not using it because this problem would come up in a class prior to that theorem being introduced
How is y^2=e in 15:56? Any hints in attacking this problem? I can only come up that y^2 is in the subgroup but cannot show that it is the identity.
I think we can use the same technique that he used in 18:30, but instead yx=x^3y , we have xy=yx. :)
Just consider the quotient group, Is the identity so y has order 2 so y² Is in and he can only be e of course
What's the intuition for landing on Z_2 x Z_2 x Z_2 as a candidate for an isomorphic group? Just wondering what intuitively makes that pop out (especially for an undergrad through 1/3rd of an abstract algebra class) Thanks :)
Because Z_2 x Z_2 x Z_2 is abelian and every non-identity element has order 2, which is easy to see because 0+0 = 0 and 1+1 = 0 in Z_2 so if you add any element (i, j, k) to itself you get (i+i, j+j, k+k) = (0, 0, 0).
If you like geometrical visualisations then Z_2 x Z_2 x Z_2 is the symmetries of a cuboid whose three sidelengths are all different. This generalises the Klein Four group Z_2 x Z_2 which is the symmetries of a non-square rectangle - Z_2 x Z_2 is generated by the two reflections of the rectangle and Z_2 x Z_2 x Z_2 is generated by the three reflections of the cuboid.
Ok yeah I need to watch this again. And probably again after that.😅
Thank you
6:57 It's actually because they have the same size _and_ they are finite sets.
8:59 This is the standard method of showing a _homomorphism_ is injective. Obviously an arbitrary function doesn't necessarily have a kernel defined (at least not a _group_ kernel; there is a more broad definition of kernel which applies to categories).
I think those are both nitpicking:
- I usually only say "size" for the cardinality of a finite set, so maybe michael uses this convention too (there's no need to use big/fancy words when not necessary - same goes for "transpositions" which should just be called "swaps"), and anyway he says that the sets are finite in the very next sentence.
- because kernels aren't defined for arbitrary functions, by saying "showing something is injective by ... it's kernel" it is implicit that the "something" has a kernel, so is a homomorphism and not just an arbitrary function
@@schweinmachtbree1013 Fair enough
10:48 worth pointing out that in general every element of order 2 is its own inverse because x^2 = x(x^(-1)).