On exercise #3, it's just to *find* an isomorphism? I tried to apply the (a₂ + b₂ = a₁ + h + b₁ + j) equality but got stuck because ℤ/nℤ consists of cosets as members of a set, and we don't seem to be talking about cosets of cosets. Or am I missing something? The isomorphism part seems pretty easy, so I feel like I might be.
I had a cool Cartesian group for #4, and everything was great until #6. My G₁ and G₂ aren't themselves groups, which made finding the kernel problematic. 😁 Back to the drawing board.
Yes, it's just to find an isomorphism. It's not a particularly difficult exercise, so you're probably overthinking something. We're just finding an isomorphism between cosets of nℤ, which look like k + nℤ, and ℤ_n, which goes from 0 to n-1.
4 through 6 are meant to be general exercises – you aren't meant to pick a specific group, but rather do it generally. (As in, you can do this for any pair of groups G₁ and G₂, so you don't need to "choose" what they are.)
A lot of meat on this bone! It's going to take me a few days to chew through this. Everything reads good at first read-through, and I'm looking forward to digging into it. After my coding binge, I'll have to do a little memory refreshing. Feels like a long time since I've done this stuff...
>> "which means that ab⁻¹ ∈ ker(ϕ). Thus, ab⁻¹ ∈ ker(ϕ), which means they are in the same right coset as well."
Are the two expressions meant to be different? The phrasing makes it sound as if the second follows from the first. I think I'm going to have to take an L on exercise #1. I'm just not getting that proof.
Those last three turned out to be kind of fun (assuming I got them right, of course). I'm still chewing on #1. I followed the proof and did a bunch of examples, but I'm not sure I *understand* the proof well enough to reverse-engineer it. Will try again later. I've spent the day updating my homework page with the new answers and lots of notes. Time for some other things today!
About to dig into the last section with the 5ℤ example. I think this is finally starting to click. I'll find out when I hit the exercises!
A minor question: In ℤ₆ are the cosets of {0,3} equally the cosets of {1,4} or {2,5}, or is it just {0,3} because it contains the identity? So, {1,4} and {2,5} would in some sense be secondary to {0,3}. Or are they equal (co-!) and the choice is arbitrary?
Wait... the color thing works for G=ℤ₆ with H={0,3} but NOT G=D₆ with H={e,s}? That surprised me until I realized the operation in D₆ isn't commutative.
>> "for example, when we take {r, s} and try to "multiply" it by itself,"
Should this be {r, rs} here and in the paragraph below?
FWIW, I pulled the red+yellow=blue image into my graphics editor and copied the yellow part over the blue part. I don't know how easy it is for you to redo it or fix it yourself, but you're welcome to it if you want it. (I'd just drop it here, but Substack doesn't support it. I can put it in a Note.)
If you look at the set of cosets of {0, 3}, they are of course the same set if you think about them instead as cosets of {1, 4} or {2, 5}. In general, you'll probably think about them in terms of whichever one contains the identity though. It makes some things easier, and we'll also see in the next installment (in the final stages of editing right now) that normal groups and kernels of homomorphisms are essentially the same thing.
The color thing working is exactly the same as saying that the subgroup is normal. And yes, in this case the reason that {e, s} is not normal in D6 is because our operation is not commutative. There are subgroups of non-commutative groups that are normal though - in fact, {1, r, r^2} is normal in D6. (I think we've seen that before somewhere, right?)
Yes, that should definitely be {r, rs}. Thanks for catching that!
I'm not sure I'm clear on exactly what it means to say "a and b are in the same coset of ker(ϕ)".
I tried two examples: First, φ: D₆ → S₃ where ker(φ) = {e}, and, as you say in the text, I get six cosets, each with one element (and left and right cosets match). Second, φ: D₆ → ℤ₂ (as in the text) where ker(φ) = {e, r, r²}, and I get the two cosets H and sH from last lesson.
So, does "same coset" refer to *any* of the cosets? (Or just the one containing the identity, is my confusion.) As an example in the second case, s and sr are in the same coset -- but not the one containing e -- and φ(s)=1 and φ(sr)=1, so φ(a)=φ(b) as required. Am I on target?
Yes, it can be any coset. Saying that an and b are in the same coset is equivalent to saying that a ker(φ) = b ker(φ), because if these sets are equal, then they are the same set. The LHS tells you a is in this set (since e in ker(φ)) and similarly, the RHS tells you b is in this set. Does that help?
I think the second case that you work out is correct, yes.
One thing I can't seem to figure out is exactly how to get from a ker(φ) = b ker(φ) to b=ak.
I can see ak=bk, where k ∈ ker(φ), but not how bk becomes b while leaving ak. Is the k in bk specifically e (while the k in ak is k ∈ ker(φ) not necessarily e)?
Let's do an example. Say that ker(φ) = {e, l, m}, and a ker(φ) = b ker(φ). What this says is that {a, al, am} = {b, bl, bm}. Since these sets are equal, we must have b = something on the left. But that means we must have b = a, b = al, or b = am. Therefore, we can write that b = ak, for some k in ker(φ).
On exercise #3, it's just to *find* an isomorphism? I tried to apply the (a₂ + b₂ = a₁ + h + b₁ + j) equality but got stuck because ℤ/nℤ consists of cosets as members of a set, and we don't seem to be talking about cosets of cosets. Or am I missing something? The isomorphism part seems pretty easy, so I feel like I might be.
I had a cool Cartesian group for #4, and everything was great until #6. My G₁ and G₂ aren't themselves groups, which made finding the kernel problematic. 😁 Back to the drawing board.
Yes, it's just to find an isomorphism. It's not a particularly difficult exercise, so you're probably overthinking something. We're just finding an isomorphism between cosets of nℤ, which look like k + nℤ, and ℤ_n, which goes from 0 to n-1.
4 through 6 are meant to be general exercises – you aren't meant to pick a specific group, but rather do it generally. (As in, you can do this for any pair of groups G₁ and G₂, so you don't need to "choose" what they are.)
Oh, okay… are #5 and #6 something of a hint in that direction, then?
Edit: Never mind. After re-reading #4, I think I see what you're asking.
A lot of meat on this bone! It's going to take me a few days to chew through this. Everything reads good at first read-through, and I'm looking forward to digging into it. After my coding binge, I'll have to do a little memory refreshing. Feels like a long time since I've done this stuff...
>> "which means that ab⁻¹ ∈ ker(ϕ). Thus, ab⁻¹ ∈ ker(ϕ), which means they are in the same right coset as well."
Are the two expressions meant to be different? The phrasing makes it sound as if the second follows from the first. I think I'm going to have to take an L on exercise #1. I'm just not getting that proof.
Those last three turned out to be kind of fun (assuming I got them right, of course). I'm still chewing on #1. I followed the proof and did a bunch of examples, but I'm not sure I *understand* the proof well enough to reverse-engineer it. Will try again later. I've spent the day updating my homework page with the new answers and lots of notes. Time for some other things today!
https://logosconcarne.substack.com/p/group-theory-notes-and-homework
About to dig into the last section with the 5ℤ example. I think this is finally starting to click. I'll find out when I hit the exercises!
A minor question: In ℤ₆ are the cosets of {0,3} equally the cosets of {1,4} or {2,5}, or is it just {0,3} because it contains the identity? So, {1,4} and {2,5} would in some sense be secondary to {0,3}. Or are they equal (co-!) and the choice is arbitrary?
Wait... the color thing works for G=ℤ₆ with H={0,3} but NOT G=D₆ with H={e,s}? That surprised me until I realized the operation in D₆ isn't commutative.
>> "for example, when we take {r, s} and try to "multiply" it by itself,"
Should this be {r, rs} here and in the paragraph below?
FWIW, I pulled the red+yellow=blue image into my graphics editor and copied the yellow part over the blue part. I don't know how easy it is for you to redo it or fix it yourself, but you're welcome to it if you want it. (I'd just drop it here, but Substack doesn't support it. I can put it in a Note.)
If you look at the set of cosets of {0, 3}, they are of course the same set if you think about them instead as cosets of {1, 4} or {2, 5}. In general, you'll probably think about them in terms of whichever one contains the identity though. It makes some things easier, and we'll also see in the next installment (in the final stages of editing right now) that normal groups and kernels of homomorphisms are essentially the same thing.
The color thing working is exactly the same as saying that the subgroup is normal. And yes, in this case the reason that {e, s} is not normal in D6 is because our operation is not commutative. There are subgroups of non-commutative groups that are normal though - in fact, {1, r, r^2} is normal in D6. (I think we've seen that before somewhere, right?)
Yes, that should definitely be {r, rs}. Thanks for catching that!
I'm not sure I'm clear on exactly what it means to say "a and b are in the same coset of ker(ϕ)".
I tried two examples: First, φ: D₆ → S₃ where ker(φ) = {e}, and, as you say in the text, I get six cosets, each with one element (and left and right cosets match). Second, φ: D₆ → ℤ₂ (as in the text) where ker(φ) = {e, r, r²}, and I get the two cosets H and sH from last lesson.
So, does "same coset" refer to *any* of the cosets? (Or just the one containing the identity, is my confusion.) As an example in the second case, s and sr are in the same coset -- but not the one containing e -- and φ(s)=1 and φ(sr)=1, so φ(a)=φ(b) as required. Am I on target?
Yes, it can be any coset. Saying that an and b are in the same coset is equivalent to saying that a ker(φ) = b ker(φ), because if these sets are equal, then they are the same set. The LHS tells you a is in this set (since e in ker(φ)) and similarly, the RHS tells you b is in this set. Does that help?
I think the second case that you work out is correct, yes.
One thing I can't seem to figure out is exactly how to get from a ker(φ) = b ker(φ) to b=ak.
I can see ak=bk, where k ∈ ker(φ), but not how bk becomes b while leaving ak. Is the k in bk specifically e (while the k in ak is k ∈ ker(φ) not necessarily e)?
Let's do an example. Say that ker(φ) = {e, l, m}, and a ker(φ) = b ker(φ). What this says is that {a, al, am} = {b, bl, bm}. Since these sets are equal, we must have b = something on the left. But that means we must have b = a, b = al, or b = am. Therefore, we can write that b = ak, for some k in ker(φ).
Does that help?
Yes! It wasn’t clear to me that ker(φ) wasn’t trivial or that x ker(φ) should be expanded to the set. Gives me a good starting point tomorrow, thanks!