I have a question about the condition (a⁻¹x ∈ H). I tried that with the exercise 2 subgroup H={e, s}, and found that multiplying them one way always gave me s (which satisfies s ∈ H), but in some cases the multiplication didn't commute, and I got, in all four exceptions, sr² (which is not in H).
Not that my answers for exercise 2 are necessarily correct, but is this result because the subgroup isn't normal, or is it sufficient that there exists an ab ∈ H?
FWIW, I've decided to start a "homework and notes" page on my blog here. I may continue to also post answers in a restack, but I may just start adding to that page. (I find the difference between Notes text and post/page text downright weird here.)
In the case of rH, which has {r, sr³} as members, there is sr³⋅r=s, and s is in H, but r⋅sr³=sr², and sr² is not in H. So, my question was whether the first case was sufficient. As mentioned in my other comment, with the two cosets of H={e, r, r², r³}, a operation between any two members results in x ∈ H.
I think what you’re seeing is that the cosets rH and Hr are not the same; we require that the cosets line up exactly to be considered the same. The issue is that, while both the left and right cosets of G partition G, unless G is normal, the partitions do not have to be the same.
They're definitely not the same. rH = {r, sr³} and Hr = {r, sr}. In both of them, ab≠ba, when multiplying the two members of the respective sets.
And r³H = {r³, sr} and Hr³ = {r³, sr³} also don't equal each other, and they are where the other case of ab≠ba is. In all four cases, the answer is sr² rather than s.
So, yeah, it seems when aH≠Ha, then it is also not the case the multiplying any two, in any order, results in x ∈ H.
Ha! So, in my answer to #3 for last time, I stumbled on the three-member cosets of D₆ without knowing it.
FWIW, the problem with x⁻¹ seems to be font related. Substack uses the "Spectral" font in the post's text, and it apparently doesn't have a good implementation of the Unicode "superscript minus" character (U+207B). Post comments use a different font entirely, so it works fine here.
I've got a full weekend so won't get to the lesson until Monday but taking a quick glance I laughed when you mentioned "arborescence". Synchronicity strikes again. I just ran into that recently in the comments on a video about an interesting maze generation algorithm.
I have a question about the condition (a⁻¹x ∈ H). I tried that with the exercise 2 subgroup H={e, s}, and found that multiplying them one way always gave me s (which satisfies s ∈ H), but in some cases the multiplication didn't commute, and I got, in all four exceptions, sr² (which is not in H).
Not that my answers for exercise 2 are necessarily correct, but is this result because the subgroup isn't normal, or is it sufficient that there exists an ab ∈ H?
FWIW, I've decided to start a "homework and notes" page on my blog here. I may continue to also post answers in a restack, but I may just start adding to that page. (I find the difference between Notes text and post/page text downright weird here.)
https://logosconcarne.substack.com/p/group-theory-notes-and-homework
It isn’t clear to me what “multiplying THEM” is referring to in your second sentence; could you expound a bit more?
OTOH, multiplying any two elements of the two cosets of H={e, r, r², r³} always results in x ∈ H.
I guess I’m also unsure what is meant by “or is it sufficient that ab in H”? Sufficient for what?
In the case of rH, which has {r, sr³} as members, there is sr³⋅r=s, and s is in H, but r⋅sr³=sr², and sr² is not in H. So, my question was whether the first case was sufficient. As mentioned in my other comment, with the two cosets of H={e, r, r², r³}, a operation between any two members results in x ∈ H.
I think what you’re seeing is that the cosets rH and Hr are not the same; we require that the cosets line up exactly to be considered the same. The issue is that, while both the left and right cosets of G partition G, unless G is normal, the partitions do not have to be the same.
They're definitely not the same. rH = {r, sr³} and Hr = {r, sr}. In both of them, ab≠ba, when multiplying the two members of the respective sets.
And r³H = {r³, sr} and Hr³ = {r³, sr³} also don't equal each other, and they are where the other case of ab≠ba is. In all four cases, the answer is sr² rather than s.
So, yeah, it seems when aH≠Ha, then it is also not the case the multiplying any two, in any order, results in x ∈ H.
Sure, here's my actual math:
H: e⋅s=s; s⋅e=s.
rH: r⋅sr³=sr²; sr³⋅r=s.
r²H: r²⋅sr²=s; sr²⋅r²=s.
r³H: r³⋅sr=sr²; sr⋅r3=s.
Hr: r⋅sr=s; sr⋅r=sr².
Hr²: r²⋅sr²=s; sr²⋅r²=s.
Hr³: r³⋅sr³=s; sr³⋅r³=sr².
In most cases, the result is s, but in four cases, it's sr².
Ha! So, in my answer to #3 for last time, I stumbled on the three-member cosets of D₆ without knowing it.
FWIW, the problem with x⁻¹ seems to be font related. Substack uses the "Spectral" font in the post's text, and it apparently doesn't have a good implementation of the Unicode "superscript minus" character (U+207B). Post comments use a different font entirely, so it works fine here.
Interesting; I’ll mess around with the fonts on the backend and see if that helps
I've got a full weekend so won't get to the lesson until Monday but taking a quick glance I laughed when you mentioned "arborescence". Synchronicity strikes again. I just ran into that recently in the comments on a video about an interesting maze generation algorithm.
Have a good weekend!