Homomorphisms and Isomorphisms -- Abstract Algebra 8

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[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.

[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!

[matheusjahnke]

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...

but 3*1 is not 1 in Z4, so it violates the injectivity that is a requirement for f being an isomorphism.

[Anonymous-cwyd]

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.

[ricardomejias]

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

[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 <φ(1)> 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).

[seanbastian]

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

[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?

[mohamedfarouk]

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

[maxgoldman]

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)).