The proof of the corollary on the board at 23:30 is not correct because in G/N you can't just replace G and N by anything they're isomorphic to - for example k *Z* is isomorphic to l *Z* for any k,l>0, but *Z* / k *Z* is certainly not isomorphic to *Z* / l *Z* since then all finite cyclic groups would be isomorphic! Also at 1:39 the equals sign should be an isomorphism sign.
Man, I salute you for spotting this. Indeed, you can't just say because N1 is isomorphic to N2 and therefore G/N1 is isomorphic to G/N2. I actually read your comment 4 times before I can understand what you're trying to say. But it's good because you lead me to think about these things more deeply, and also why the example provided by Michael works. For that, thank you. In your example, kZ is isomorphic to lZ. As long as you're operating under their respective little own worlds (i.e. kZ and lZ), k and l are interchangeable. But in forming a quotient group with their parent Z, k and l, and hence the elements in kZ and lZ, have to interact with other elements in Z outside their little world, and that's where things break down. Obviously, you can't interchange k with l anymore. The actual numbers that k and l represent matter. Put it in another way, "treat k as if it's an l and vice versa" does not make sense when you extend it to the entire set of integers. On the other hand, let's take a look at Michael's example. Zn is not only isomorphic to Z/nZ but they're indeed the same group. On the surface, Zn seems to be just the set {0, 1, 2, ... (n-1) } but in fact they represent the (n-1) congruent classes mod n: [0], [1], [2], ... [n-1] which are sets themselves. Here's Zn: {...,-2n, -n, 0, n, 2n, ...} {...,-2n+1, -n+1, 1, n+1, 2n+1, ...} {...,-2n+2, -n+2, 2, n+2, 2n+2, ...} ... {...,-2n+(n-1), -n+(n-1), n-1, n+(n-1), 2n+(n-1),...} And these are the same elements you'll find in Z/nZ. Similarly, is not simply {0, d, 2d, 3d, ... (n/d-1)*d} but congruent classes mod n. They are also sets themselves. So is not just isomorphic to dZ/nZ, they're the same group indeed. Consequently, Michael's example works because he's not replacing G and N with isomorphic groups but with the same group. In other words, he should have said that Zn/ equals (Z/nZ) / (dZ/nZ) instead of Zn/ is isomorphic to (Z/nZ) / (dZ/nZ).
@@llchan Thank you so much for this comment, this cleared some things up for me too. I'm glad these comments fill the gaps that are sometimes missing in the videos.
@ 30:08 you said "in fact we've shown that all subgroups of G/N are of the form H/N". Is that because from [g ∈ H ⇔ gN ∈ K] in the purple box we have [gN ∈ H/N ⇔ gN ∈ K], so K=H/N?
@8:48 wouldn’t we need double inclusion? So we have kernel is a subset of H intersect N, but we need to show that H intersect N is a subset of the kernel.
I think Michael’s presentation is not clear and it leads to your question. Here’s how I would present it so that your question goes away. We’re trying to find elements in H such that phi(h)=N. We arrive at the conclusion that the h’s that satisfy hN = N are the elements we’re looking for. Any element in the set H intersect N satisfies the condition. Therefore kernel of phi is the set H intersect N. In essence, we are not trying to prove that 2 sets are equal because we are not given 2 sets in the first place. We’re trying to find a set that satisfies certain requirements. Therefore, we don’t need double inclusion. His presentation confused you into thinking that we’re trying to prove that a certain set is equal to another set.
For the second isomorphism theorem: Don't we have to proof first that N is a normal subgroup of HN, so that the quotient group HN/N is even well-defined? Also, in the third theorem, don't we have to require that N is a _normal_ subgroup of M? 30:20 We didn't show that N is a subgroup of H - is that somehow obvious? Don't we even have to show that N is a _normal_ subgroup of H, so that the quotient group H/N is well-defined?
(1) Yes, N must be a normal subgroup of HN for HN/N to make sense. Since N is already a normal subgroup of G, it suffices to show that N is a subgroup of HN. HN consists of elements of H times elements of N. Pick e from H and multiply it through all elements of N and you’ll get N and therefore HN contains N. In other words, N is a subset of HN. Since N is a normal subgroup, it must have already satisfied the 4 requirements to be a group (closure, associative, existence of identity, existence of inverse) and so N is indeed a subgroup of HN. Ah, I guess HN also needs to be a subgroup of G but I think that’s quite easy to show. Without that, you can’t use the word subgroup and you can only say that the group N is just contained in a larger set HN. (2) Yes, N has to be a normal subgroup of M. But we already assume N to be a normal subgroup of G, and N a subgroup of M. It follows that N is a normal subgroup of M because M is a subgroup of G. In other words, we know that g*n*g^(-1) is in N for every g in G and so it is automatically true for every g in M, a subset of G. (3) Yes, we need N to be a normal subgroup of H. I wouldn’t call it obvious. H is constructed from K by including all the elements of its cosets. Since K contains the coset eN=N, H has all the elements of N => N is a subset of H. Similar to the argument in (1) above, N is a subgroup of H (that is, N is already a group and N is a subset of a group H). Given our premise that N is a normal subgroup of G, N must be a normal subgroup of H (see (2) above). Thanks to your question because it makes me think about these proofs deeply. Before answering your question, I made the mistake in thinking that if N is a normal subgroup of G, it must also be a normal subgroup of every subgroup of G. That is NOT true. Suppose N is a normal subgroup of G, and H is a subgroup of G. To show that N is a normal subgroup of H, we still have to show that N is a subgroup of H.
you might want to (at least as far as I can tell) want to rewrite the x(y^-1) you got in the well defined proof to be (x^-1)y. I cant see why those would be analogous and the later is what I and you proved in the coset video.
You should make a lecture about proofs. Like.. what is mathematical proof. Why does a proof actually prove that what was said is demonstrable from the hypothesis...
If you spell everything out in absolutely complete detail, mathematical proofs use nothing more than the primitive logical notions of "and", "or", "not", "implies", "for all", "for some", some notion of collections (the most common choice is sets), and the chosen axioms for the things being studied. For example if you want to prove results about whole numbers then the standard system of axioms is called the Peano axioms, and consist of: - x + 0 = x - x + (y+1) = (x+y) + 1 - x × 0 = 0 - x × (y+1) = x×y + x - x+1 = y+1 ⇒ x=y - there is no (non-negative) whole number n such that n+1 = 0 - for a family of statements P(n), if P(0) is true and P(k) implies P(k+1) for all k, then P(n) is true for all n (that is, if P(0) is true and P(0) ⇒ P(1) ⇒ P(2) ⇒ P(3) ⇒ ... then P(n) is true for all n). These latter two axioms can be written in more formal language as "¬∃n:(n+1=0)" and "(P(0) & ∀k:(P(k)⇒P(k+1))) ⇒ ∀n:P(n)". For something more sophisticated like groups, the standard axioms are: - (g ∘ h) ∘ k = g ∘ (h ∘ k) - e_G ∘ g = g = g ∘ e_G - g ∘ g^(-1) = e_G = g^(-1) ∘ g. Because formal mathematical proofs are so rigid at the base level (i.e. when completely spelled out), one doesn't even have to know what the symbols mean, just that each step follows from the previous ones using the axioms and basic logic, and in this sense mathematical proofs can be seen as nothing more than a game of moving symbols around subject to the given rules (axioms). Of course for mathematicians understanding the symbols and the concepts is the most important thing, because you're never going to get anywhere by blindly guessing, but if we want to _check_ a proof, perhaps an extremely long or technical one, a computer can easily be taught the rules and can check in an instant whether a formal proof follows them. We are at the point now where all of undergraduate mathematics has been completely formalised in this way and checked by proof-checkers, and these days we have not only proof-checkers but also proof-assistants which can suggest possible next steps in a proof (my favourite being Lean), and even some automated-theorem-provers which attempt to prove results all by themselves; after all, blindly guessing isn't that bad if you can make thousands of blind guesses every second.
The proof of the corollary on the board at 23:30 is not correct because in G/N you can't just replace G and N by anything they're isomorphic to - for example k *Z* is isomorphic to l *Z* for any k,l>0, but *Z* / k *Z* is certainly not isomorphic to *Z* / l *Z* since then all finite cyclic groups would be isomorphic!
Also at 1:39 the equals sign should be an isomorphism sign.
Man, I salute you for spotting this. Indeed, you can't just say because N1 is isomorphic to N2 and therefore G/N1 is isomorphic to G/N2. I actually read your comment 4 times before I can understand what you're trying to say. But it's good because you lead me to think about these things more deeply, and also why the example provided by Michael works. For that, thank you.
In your example, kZ is isomorphic to lZ. As long as you're operating under their respective little own worlds (i.e. kZ and lZ), k and l are interchangeable. But in forming a quotient group with their parent Z, k and l, and hence the elements in kZ and lZ, have to interact with other elements in Z outside their little world, and that's where things break down. Obviously, you can't interchange k with l anymore. The actual numbers that k and l represent matter. Put it in another way, "treat k as if it's an l and vice versa" does not make sense when you extend it to the entire set of integers.
On the other hand, let's take a look at Michael's example.
Zn is not only isomorphic to Z/nZ but they're indeed the same group. On the surface, Zn seems to be just the set {0, 1, 2, ... (n-1) } but in fact they represent the (n-1) congruent classes mod n: [0], [1], [2], ... [n-1] which are sets themselves. Here's Zn:
{...,-2n, -n, 0, n, 2n, ...}
{...,-2n+1, -n+1, 1, n+1, 2n+1, ...}
{...,-2n+2, -n+2, 2, n+2, 2n+2, ...}
...
{...,-2n+(n-1), -n+(n-1), n-1, n+(n-1), 2n+(n-1),...}
And these are the same elements you'll find in Z/nZ.
Similarly, is not simply {0, d, 2d, 3d, ... (n/d-1)*d} but congruent classes mod n. They are also sets themselves. So is not just isomorphic to dZ/nZ, they're the same group indeed.
Consequently, Michael's example works because he's not replacing G and N with isomorphic groups but with the same group. In other words, he should have said that Zn/ equals (Z/nZ) / (dZ/nZ) instead of Zn/ is isomorphic to (Z/nZ) / (dZ/nZ).
@@llchan Thank you so much for this comment, this cleared some things up for me too. I'm glad these comments fill the gaps that are sometimes missing in the videos.
@ 30:08 you said "in fact we've shown that all subgroups of G/N are of the form H/N". Is that because from [g ∈ H ⇔ gN ∈ K] in the purple box we have [gN ∈ H/N ⇔ gN ∈ K], so K=H/N?
@8:48 wouldn’t we need double inclusion?
So we have kernel is a subset of H intersect N, but we need to show that H intersect N is a subset of the kernel.
Same question @20:04
@@Happy_Abe By the iff's double implication.
@@r.maelstrom4810 thank you!
I think Michael’s presentation is not clear and it leads to your question. Here’s how I would present it so that your question goes away.
We’re trying to find elements in H such that phi(h)=N. We arrive at the conclusion that the h’s that satisfy hN = N are the elements we’re looking for. Any element in the set H intersect N satisfies the condition. Therefore kernel of phi is the set H intersect N.
In essence, we are not trying to prove that 2 sets are equal because we are not given 2 sets in the first place. We’re trying to find a set that satisfies certain requirements. Therefore, we don’t need double inclusion. His presentation confused you into thinking that we’re trying to prove that a certain set is equal to another set.
For the second isomorphism theorem: Don't we have to proof first that N is a normal subgroup of HN, so that the quotient group HN/N is even well-defined?
Also, in the third theorem, don't we have to require that N is a _normal_ subgroup of M?
30:20 We didn't show that N is a subgroup of H - is that somehow obvious? Don't we even have to show that N is a _normal_ subgroup of H, so that the quotient group H/N is well-defined?
(1) Yes, N must be a normal subgroup of HN for HN/N to make sense. Since N is already a normal subgroup of G, it suffices to show that N is a subgroup of HN. HN consists of elements of H times elements of N. Pick e from H and multiply it through all elements of N and you’ll get N and therefore HN contains N. In other words, N is a subset of HN. Since N is a normal subgroup, it must have already satisfied the 4 requirements to be a group (closure, associative, existence of identity, existence of inverse) and so N is indeed a subgroup of HN. Ah, I guess HN also needs to be a subgroup of G but I think that’s quite easy to show. Without that, you can’t use the word subgroup and you can only say that the group N is just contained in a larger set HN.
(2) Yes, N has to be a normal subgroup of M. But we already assume N to be a normal subgroup of G, and N a subgroup of M. It follows that N is a normal subgroup of M because M is a subgroup of G. In other words, we know that g*n*g^(-1) is in N for every g in G and so it is automatically true for every g in M, a subset of G.
(3) Yes, we need N to be a normal subgroup of H. I wouldn’t call it obvious. H is constructed from K by including all the elements of its cosets. Since K contains the coset eN=N, H has all the elements of N => N is a subset of H. Similar to the argument in (1) above, N is a subgroup of H (that is, N is already a group and N is a subset of a group H). Given our premise that N is a normal subgroup of G, N must be a normal subgroup of H (see (2) above).
Thanks to your question because it makes me think about these proofs deeply. Before answering your question, I made the mistake in thinking that if N is a normal subgroup of G, it must also be a normal subgroup of every subgroup of G. That is NOT true. Suppose N is a normal subgroup of G, and H is a subgroup of G. To show that N is a normal subgroup of H, we still have to show that N is a subgroup of H.
@@llchan Thanks for the long reply. :)
you might want to (at least as far as I can tell) want to rewrite the x(y^-1) you got in the well defined proof to be (x^-1)y. I cant see why those would be analogous and the later is what I and you proved in the coset video.
You should make a lecture about proofs. Like.. what is mathematical proof. Why does a proof actually prove that what was said is demonstrable from the hypothesis...
Look here: www.youtube.com/@mathmajor/playlists
If you spell everything out in absolutely complete detail, mathematical proofs use nothing more than the primitive logical notions of "and", "or", "not", "implies", "for all", "for some", some notion of collections (the most common choice is sets), and the chosen axioms for the things being studied.
For example if you want to prove results about whole numbers then the standard system of axioms is called the Peano axioms, and consist of:
- x + 0 = x
- x + (y+1) = (x+y) + 1
- x × 0 = 0
- x × (y+1) = x×y + x
- x+1 = y+1 ⇒ x=y
- there is no (non-negative) whole number n such that n+1 = 0
- for a family of statements P(n), if P(0) is true and P(k) implies P(k+1) for all k, then P(n) is true for all n (that is, if P(0) is true and P(0) ⇒ P(1) ⇒ P(2) ⇒ P(3) ⇒ ... then P(n) is true for all n).
These latter two axioms can be written in more formal language as "¬∃n:(n+1=0)" and "(P(0) & ∀k:(P(k)⇒P(k+1))) ⇒ ∀n:P(n)". For something more sophisticated like groups, the standard axioms are:
- (g ∘ h) ∘ k = g ∘ (h ∘ k)
- e_G ∘ g = g = g ∘ e_G
- g ∘ g^(-1) = e_G = g^(-1) ∘ g.
Because formal mathematical proofs are so rigid at the base level (i.e. when completely spelled out), one doesn't even have to know what the symbols mean, just that each step follows from the previous ones using the axioms and basic logic, and in this sense mathematical proofs can be seen as nothing more than a game of moving symbols around subject to the given rules (axioms). Of course for mathematicians understanding the symbols and the concepts is the most important thing, because you're never going to get anywhere by blindly guessing, but if we want to _check_ a proof, perhaps an extremely long or technical one, a computer can easily be taught the rules and can check in an instant whether a formal proof follows them. We are at the point now where all of undergraduate mathematics has been completely formalised in this way and checked by proof-checkers, and these days we have not only proof-checkers but also proof-assistants which can suggest possible next steps in a proof (my favourite being Lean), and even some automated-theorem-provers which attempt to prove results all by themselves; after all, blindly guessing isn't that bad if you can make thousands of blind guesses every second.