Homomorphisms and Isomorphisms -- Abstract Algebra 8

[@cabritoguitarrista]

Hey, I just want to say that i really appreciate this you're doing here on youtube. I'm currently doing a bachelor on electrical engineering, but what I really wanted to do is math, so it's great to be able to learn all this interesting stuff online. Thank you for not letting my dream die, sir!

[@matematicacommarcospaulo]

Does your course abstract Algebra?

[@cabritoguitarrista]

​@@matematicacommarcospaulo It doesn't, that's why i'm so grateful to be able to learn it here

[@matematicacommarcospaulo]

I have two videos about rings

[@M.athematech]

regarding r_g(x), of course defining it as xg still gives you a permutation, the issue is that the assigment g |-> r_g would not be a homomorphism but an anti-homorphism not that xg isn't a permutation

[@SeeTv.]

Great video as always! 15:21 I have another proof of this fact which highlights the different structure of the additive and multiplicative group quite nicely: Note that in the multiplicative group there IS an element of order 2, namely -1, i.e. (-1)^2 = 1. (The solutions to x^2 = 1 are 1 and -1, but 1 is the identity and has order 1) In the additive group however, there is no element of order 2, because x+x = 0 2x = 0 x = 0, but 0 is the identity with order 1. Indeed, by the same argument (with n instead of 2), each element in the additive group has order infinity, except the identity, which has order 1 (so there are no elements whose order is a natural number ≥ 2). So intuitively, the two groups have a different structure. Now for the proof to make this intuition more precise: Assume there was an isomorphism f from the additive group of rationals to the multiplicative group of non-zero rationals. Note that every homomorphism sends identities to identities (by 37:43), so here we have f(0) = 1. Since f is an isomorphism, it is onto, so there is an a in the additive rationals which gets mapped to -1, i.e. f(a) = -1. This element a cannot be 0, because f is a function which already maps 0 to 1. Now observe that (-1)^2 = 1 (-1 has order 2), so (f(a))^2 = 1. By the homomorphism property we get 1 = (f(a))^2 = f(a)f(a) = f(a+a) = f(2a). But since f is injective we have the implication 1= f(2a) = f(0) => 2a = 0 => a = 0, which is a contradiction because as noted above a (which is mapped to -1) cannot be 0. QED This proof is also applicable to the case where you exchange Q with R: The additive group of real numbers is not isomprohic to the multiplicative group of non-zero real numbers. (Also in R -1 has order 2, but in the additive group no element has this property). However, the additive group of real numbers is isomorphic to the multiplicative group of POSITIVE real numbers (strictly greater than 0) via the exponential map. So by restricting the multiplicative group to the positives we get an isomorphism. The proof in this video is still very nice because the same argument also shows that the above observation is not the case for rationals. If you try to make an isomorphism from the additive group of rationals to the multiplicative group of POSITIVE rationals you get the same contradiction (the only difference to the video is that here you can ignore the negative sqrt(2) case). What I love about abstract algebra is that there are often so many different and insightful ways to prove things.

[@arjunsinha4015]

It's Morphin time !!!!

[@aashsyed1277]

Oh yeah 👍

[@matematicacommarcospaulo]

😅 I prefer the Changemen

[@Happy_Abe]

Best use of this joke 😂

[@matheusjahnke8643]

30:08 As you said, U8 is not isomorphic to Z4 even though they both have four elements. Essentially, they don't have the same structure. 1²=1=1+8n 3²=9=1+8n 5²=25=1+8n 7²=49=1+8n This means for whenever element x in U8, we have x²=e This doesn't happen in Z4, here there are only 2 elements x such that 2x=[0]... [0] and [2] for [1] and [3], they generate Z4. Let's say there was a isomorphism f between Z4 and U8 Consider 1... f(3*1)=f(1)f(1)f(1)=(f(1))²=f(1) but 3*1 is not 1 in Z4, so it violates the injectivity that is a requirement for f being an isomorphism.

[@Anonymous-cw4yd]

30:16 I think, you should have first shown H={l_g such that g is in G} is a subgroup of S_G. So saying phi is a homomorphism make sense.

[@schweinmachtbree1013]

You're right, strictly speaking the theorem at 30:16 uses the result on the next board at 42:39 (the fifth bullet point), and I think it's very unusual to prove Cayley's theorem (that every group is isomorphic to a subgroup of a symmetric group) before establishing basic properties of homomorphisms. However it is possible to fix the incorrect ordering of the results if you phrase Cayley's theorem as "every group embeds into a symmetric group", where "G_1 embeds into G_2" means that there is an injective homomorphism from G_1 into G_2 (again I'm not saying that one _should_ present things this way; only that it is possible and fixes the logical problem with the ordering). I'll also mention, because it's bugging me, that using the fact that G is non-empty at 30:33 is unneccesary - S_X is a group even if X is empty. The order of S_X is (let's say X is finite for simplicity) |X|!, and this holds even for |X|=0 - it just means that S_0 and S_1 are isomorphic symmetric groups, corresponding to the fact that 0! = 1!

[@FelipeMontealegreS]

This man is an American hero

[@mohamedfarouk9654]

32:27 I don't get why can't we right multiply by g?

[@Happy_Abe]

I have the same question if anyone knows

[@schweinmachtbree1013]

@@Happy_Abe Mathematech answered this in another comment: "regarding r_g(x), of course defining it as xg still gives you a permutation, the issue is that the assignment g ↦ r_g would not be a homomorphism (but rather an anti-homomorphism), not that xg isn't a permutation"

[@Happy_Abe]

@@schweinmachtbree1013 that makes sense but not what Michael spas in the video. He said you don’t get a permutation from it. So maybe that was a mistake

[@schweinmachtbree1013]

@@Happy_Abe Yes it was a mistake - it is easy to check that x ↦ xg is a permutation.

[@Happy_Abe]

@@schweinmachtbree1013 thought so, thus my confusion, thanks!

[@davidemasi__]

53:05 all the homomorphisms follow φ(1) = 0, φ(1) = 6 and [φ(1) = 3 or φ(1) = 9]. In fact, as you proved in the video, ord[φ(n)] | ord(n), which means ord[φ(1)] | ord(1) = 8. So there are 4 possibilities for ord[φ(1)]: it can either be 1, 2, 4 or 8. Note that ord[φ(1)] = 8 would contradict the Lagrange's theorem, since is a subgroup of Z_12 and thus ord[φ(1)] | 12. Also observe that 8 = 0 (mod 8), so φ(8) = φ(0) = 0 and this shows, in another way, why it can't be φ(1) = 2 or φ(1) = 4 as one could imagine: in the first case you would have φ(8) = 8 φ(1) = 8 * 2 = 16 = 4 (mod 12) ≠ 0 = φ(0); in the second one φ(8) = 8 φ(1) = 8 * 4 = 32 = 8 (mod 12) ≠ 0 = φ(0).

[@GugasSoares]

this is the best topic

[@ricardomejias156]

Michael, what books you recommend for love follow studying abstract algebra?

[@Happy_Abe]

How would one do the second warm up problem? Does it have to do with the order of a homomorphism of an element dividing the order of that element?

[@richardschreier3866]

ord(g₂) divides ord(g₁) = 8. Also ord(g₂) divides ord(Z₁₂) = 12. Consequently ord(g₂) = 1, 2, or 4, which are satisfied for g₂= 0, 6, or (3 or 9), respectively. Michael does a similar problem in the Examples 8 video.

[@Happy_Abe]

@@richardschreier3866 thanks

[@AbuMaxime]

1 is order 8 in Z8, therefore the order of the image of 1 in Z12 by a homomorphism must divide 8. The image of 1, therefore, can have order 1, 2, 4, or 8. (8 will have to be excluded because it doesn't divide 12). Say you look for solutions of order 2. In Z12, the only possibility is phi(1) = 6. You get all other mappings using phi(1^k) = 6^k, or actually 6xk using additive notation.

[@Happy_Abe]

@@AbuMaximethanks!

[@seanbastian4614]

What is Z^2? Is that the set {0,1} or is that the squared integers?

[@SeeTv.]

It's the set of orderes pairs of integers (just like R^2 is the plane, i.e. the set of ordered pairs of real numbers). It's also clear by the definition of the map: The input is (m,n) - an orderes pair of integers. This is standard notation for the cartesian product of sets. You never write Z^2 for the set of square numbers, that would be inconsistent.

[@maxgoldman8903]

To prove ord(g) = ord (phi(g)) at 52:00, I think we should start with g^m: Let ord(phi(g))=m. So G2=< e2, phi(g), (phi(g))^2, …(phi(g))^m-1>. g^m = phi^(-1)(phi(g^m)) = phi^(-1)(phi(g)…phi(g)) (since phi is an isomorphism, i.e. homomorphism) = phi^(-1)(phi(g))^m) = phi^(-1)(e2) = e1 (since phi(e1)=e2) If ord(g)=n, it implies n | m (namely, ord(g) | ord(phi(g)) ). Together with the previous result ord(phi(g)) | ord(g), ord(g) = ord(phi(g)).

[@schweinmachtbree1013]

ord(phi(g)) = m does not mean that G_2 = ; an arbitrary element (here phi(g) of G_2) usually does not generate the group - when it does, the group is a cyclic group (with that element as a generator), but most groups are not cyclic. There's nothing wrong with Michael's proof of 52:00 - it is the "obvious" way to do it (we already know that ord(phi(g)) | ord(g) for homomorphisms, so since phi^(-1) is also a homomorphism we get the other divisibility condition ord(g) | ord(phi(g)) that we need straight away, by writing ord(g) as ord( phi^(-1)(phi(g)) ).)

[@maxgoldman8903]

My bad. You’re indeed correct. Thank you. 😊