EDIT: ghg^(-1) being inside H does not automatically imply gHg^(-1) = H, but we also need the other way around, i.e. for all h in H, h can be expressed gh*g^(-1) for some h* in H. There isn't any issue with that if we are proving that H is a normal subgroup though (think about why). If you want to spare a few moments to help this channel, log your math levels on the Google form: forms.gle/QJ29hocF9uQAyZyH6 This helps me create content more geared towards viewers like you as I can adjust the topic of the videos, pacing, etc. Subscribe with notifications on if you like this group theory series and share this video if you think it might be useful for others learning group theory as well!
Can you elaborate on why? I tried to think of it and I can't understand why, can you give an example where all ghg^1 are in H for all h of H, but gHg^-1 is only a subset of H?
@@orangus01 This puzzled me as well.. one possibility might be gHg^(-1) reduces every h in H to the identity or a single symmetry? I must confess that I have no idea what kind of shape might result in what I just said... I was thinking somewhere along the lines of an eigenvector reducing the rank of a matrix.. perhaps there exists a conjugation that reduces the 'cardinality' of the group Sorry I have no formal experience with Group Theory at all, but am genuinely curious as well
@@wgllgw I don't think that argument works, [g] is a symmetry and thus should be reversible, meaning gHg^-1 should be a bijection to itself (a permutation), right?
I first came across this concept when solving rubik's cubes. Sometimes you have a sequence of moves that permutes 3 pieces, but not the 3 pieces that you want to permutes. So you do a setup, sequence, undo setup. This is a conjugation. Then, when I was watching 3b1b's video about diagonalization of a matrix, I realized it's the exact same concept, P^-1 DP. You can also use this to rotate the complex plane around any given point by translating, rotating, undoing translation. I love how group theory brings together all of this.
I love this series! I found 3b1b's essence of linear algebra really useful and i'm so happy someone else is using this kind of format on more arguments. Keep up the good work
I know, this might be a dumb question, but can anyone tell me, why anticlockwise rotation is same as clockwise rotation after changing perspective. Thanks in advance!
Not a dumb question at all! Are you referring to the case where I showed srs^(-1)=r^(-1) using only changing perspectives? Well, the change of perspective now is s, the horizontal reflection, so srs^(-1) is actually an anticlockwise rotation viewed in a horizontally reflected perspective, kind of like putting a mirror at the horizontal axis. If you are doing an anticlockwise rotation, then *in the mirror*, you are doing a clockwise rotation! The /looking in the mirror/ part is just /changing perspectives/ here.
@@mathemaniac i am sorry, but I am confused about the whole mirror thing, what does put a mirror at the horizontal line mean, it means I watched the image inside it instead of real life ? I cracked my head and still dont know what to look at
@@mathemaniac This is a very useful tip with the mirror. So the mirror works for understanding rotations in a horizontally reflected perspective, i.e. s *and* s^(-1)? Is there a similar trick for rsr(-1)?
@@mathemaniac I think I get the whole changing perspectives thing but i am a little confused with how when you do the final reflection s it dosent just negate the changed perspective point back to its original location where there was no conjugations
At 9:32, what is the element g, explictly? Under such symmetry g, the group has changed completely (in fact it now acts on a square instead of a cube). So I don't understand how it would make sense to even compare the subgroups H and gHg^{-1}, since they seem to be subgroups of different groups. I know how to algebraically prove that H is not normal in G, but I don't understand the intuition behind this example. Thanks for anyone who would like to help me.
It isn't actually acting on a square, it is still acting on a cube just viewing from the top! The axis goes right through the screen, so it can't be shown, and it just *seems* that gHg^(-1) is a group of symmetries of a square. The element g could be a 90 degree rotation along the third axis (apart from the two described by H and gHg^(-1))
This confuses me too. When you say an action has to preserve the object, doesn’t that include it’s orientation (which seems to have been changed in this case). Or else, wouldn’t “rotate by 1 degree” be a valid group action?
There is a distinction to be made between the group itself and group actions. The cubes and squares are for visualization. You can either understand the squares are just top views of the cube OR that the group is acting on 8 vertices and the squares show the action restricted to the top 4 vertices.
Normal subgroups? More like "No bad videos, yep!" Thanks again for putting together such a wonderful series! You really do explain so many things so much better than anything I've come across before.
this video really helped me a lot! I looked at the proof for the second isomorphism theorem here: ua-cam.com/video/2Y-nsro1bBo/v-deo.html and when g' was introduced, I needed to pause for a while and think about what that meant, but after watching this video and going back to the proof, I only needed to pause for 3 seconds before I was like, "well, obviously"... So thank you very much!
Yes, this is true. An error on my part. UA-cam doesn't allow reuploading just to correct a small error, so I am forced to leave it uncorrected here, but I will update the pinned comment to warn people of the mistake. Thanks for pointing it out.
I think the most intuitive way to understand what a normal subgroup is is to think of them as a subgroup H where H acting on elements of g is the same as elements of g acting on H.
Fuck me, I needed this when I was studying for my algebra course last year haha, I guess it doesn’t hurt to revise though. I was actually looking for stuff like this very hard on the internet, but I couldn’t obviously find them because well, they weren’t there yet
Thanks Mathemaniac, this clarifies a lot. One question: S3 (group of all symmetries of the triangle), how can we find its (normal) subgroup(s)? By Lagrange, the order of |H| \in {1, 2, 3, 6}, with {e} and {e, r, r2, f, rf, r2f} have orders 1 and 6. But for other subsets, e.g. {e, r, f}, why is this (not) a subgroup and why is (not) it normal?
My previous chapter about Lagrange's theorem said something about subgroups: the set has to be closed under composition, e.g. rf is not inside {e, r, f}, and so this is not a subgroup, let alone a normal one. {e, f} on the other hand is a subgroup, but not normal. Composing reflection with itself is the identity, and so this is a subgroup, but as discussed in the video, viewing f in a rotated perspective, i.e. rfr^(-1), is not e or f (but should be r^2 f), so this subgroup is not normal.
@@mathemaniac Is {e, r^2f} is a subgroup? Composing r^2f with itself gives r^2f*r^2f, but I don't see how this is either e or r^2f... If I would check if it's normal, I'd for example do r r^2f r^(-1)?
You might have to learn more about how these elements are related to each other, specifically rs = sr^(-1). That might help. The other way would be to recognize that f, rf, r^2 f are reflections along different axes. So a composition of two reflections would be e, and so {e, r^2 f} is a subgroup, but not normal. Yes, you can definitely check the subgroup is not normal by actually doing r (r^2 f) r^(-1), but the point of this video is that you can see it directly by thinking of these as symmetries.
@@mathemaniac Thanks. Yes, rs = sr^(-1) is true. if I compose f and rf , the this is not equal to e. If I call the vertices in a triangle {A, B, C} and I first reflect vertex A followed vertex B, then I get {A, C, B} followed by {B, C, A}. What am I missing? Or should I somehow shift the perspective like in 3:35?
EDIT: ghg^(-1) being inside H does not automatically imply gHg^(-1) = H, but we also need the other way around, i.e. for all h in H, h can be expressed gh*g^(-1) for some h* in H. There isn't any issue with that if we are proving that H is a normal subgroup though (think about why).
If you want to spare a few moments to help this channel, log your math levels on the Google form: forms.gle/QJ29hocF9uQAyZyH6
This helps me create content more geared towards viewers like you as I can adjust the topic of the videos, pacing, etc.
Subscribe with notifications on if you like this group theory series and share this video if you think it might be useful for others learning group theory as well!
Can you elaborate on why? I tried to think of it and I can't understand why, can you give an example where all ghg^1 are in H for all h of H, but gHg^-1 is only a subset of H?
@@orangus01 This puzzled me as well.. one possibility might be gHg^(-1) reduces every h in H to the identity or a single symmetry? I must confess that I have no idea what kind of shape might result in what I just said...
I was thinking somewhere along the lines of an eigenvector reducing the rank of a matrix.. perhaps there exists a conjugation that reduces the 'cardinality' of the group
Sorry I have no formal experience with Group Theory at all, but am genuinely curious as well
@@wgllgw I don't think that argument works, [g] is a symmetry and thus should be reversible, meaning gHg^-1 should be a bijection to itself (a permutation), right?
I first came across this concept when solving rubik's cubes. Sometimes you have a sequence of moves that permutes 3 pieces, but not the 3 pieces that you want to permutes. So you do a setup, sequence, undo setup. This is a conjugation. Then, when I was watching 3b1b's video about diagonalization of a matrix, I realized it's the exact same concept, P^-1 DP. You can also use this to rotate the complex plane around any given point by translating, rotating, undoing translation. I love how group theory brings together all of this.
I love this series!
I found 3b1b's essence of linear algebra really useful and i'm so happy someone else is using this kind of format on more arguments.
Keep up the good work
Thanks for the kind words!
You’re good. Great animations. I’d focus more on motivation, narrative arcs. Just an idea. Keep going.
Thanks! The idea is noted!
4:19 Put label / number on vertex hexagon to illustrate the conjiugaison etc...
This helped me finalize my understanding of geometric algebra. Thank you from the bottom of my heart.
Glad it helps!
I know, this might be a dumb question, but can anyone tell me, why anticlockwise rotation is same as clockwise rotation after changing perspective. Thanks in advance!
Not a dumb question at all! Are you referring to the case where I showed srs^(-1)=r^(-1) using only changing perspectives? Well, the change of perspective now is s, the horizontal reflection, so srs^(-1) is actually an anticlockwise rotation viewed in a horizontally reflected perspective, kind of like putting a mirror at the horizontal axis. If you are doing an anticlockwise rotation, then *in the mirror*, you are doing a clockwise rotation! The /looking in the mirror/ part is just /changing perspectives/ here.
@@mathemaniac i am sorry, but I am confused about the whole mirror thing, what does put a mirror at the horizontal line mean, it means I watched the image inside it instead of real life ? I cracked my head and still dont know what to look at
@@tramquangpho Yes - you watch the image inside the mirror!
@@mathemaniac This is a very useful tip with the mirror. So the mirror works for understanding rotations in a horizontally reflected perspective, i.e. s *and* s^(-1)?
Is there a similar trick for rsr(-1)?
@@mathemaniac I think I get the whole changing perspectives thing but i am a little confused with how when you do the final reflection s it dosent just negate the changed perspective point back to its original location where there was no conjugations
this was very helpful, THANK YOU!
Glad to help!
At 9:32, what is the element g, explictly? Under such symmetry g, the group has changed completely (in fact it now acts on a square instead of a cube). So I don't understand how it would make sense to even compare the subgroups H and gHg^{-1}, since they seem to be subgroups of different groups. I know how to algebraically prove that H is not normal in G, but I don't understand the intuition behind this example. Thanks for anyone who would like to help me.
It isn't actually acting on a square, it is still acting on a cube just viewing from the top! The axis goes right through the screen, so it can't be shown, and it just *seems* that gHg^(-1) is a group of symmetries of a square.
The element g could be a 90 degree rotation along the third axis (apart from the two described by H and gHg^(-1))
This confuses me too. When you say an action has to preserve the object, doesn’t that include it’s orientation (which seems to have been changed in this case). Or else, wouldn’t “rotate by 1 degree” be a valid group action?
There is a distinction to be made between the group itself and group actions. The cubes and squares are for visualization. You can either understand the squares are just top views of the cube OR that the group is acting on 8 vertices and the squares show the action restricted to the top 4 vertices.
Wonderful geometric insights beyond the algebriac structural relationships.
Normal subgroups? More like "No bad videos, yep!" Thanks again for putting together such a wonderful series! You really do explain so many things so much better than anything I've come across before.
this video really helped me a lot! I looked at the proof for the second isomorphism theorem here: ua-cam.com/video/2Y-nsro1bBo/v-deo.html and when g' was introduced, I needed to pause for a while and think about what that meant, but after watching this video and going back to the proof, I only needed to pause for 3 seconds before I was like, "well, obviously"... So thank you very much!
Hey, thank you for this series. However, I don't quite see why ghg^{-1} \in H \implies gHg^{-1}=H. Equality does not follow trivially, only subset.
Yes, this is true. An error on my part. UA-cam doesn't allow reuploading just to correct a small error, so I am forced to leave it uncorrected here, but I will update the pinned comment to warn people of the mistake. Thanks for pointing it out.
I need to develop more intuition! lol
I got a very square mindset after so many calculus jajaja
The big problems in mathematics is lack of clarity
Very great!! Many thanks for the video
Glad to hear that you enjoyed the video!
The point you make around the 4:30 to 5:15 mark is very helpful. Definitely worth you repeating it. Thank you.
Dude, THANK YOU!!!! For these videos
what a nice video
I think the most intuitive way to understand what a normal subgroup is is to think of them as a subgroup H where H acting on elements of g is the same as elements of g acting on H.
Fuck me, I needed this when I was studying for my algebra course last year haha, I guess it doesn’t hurt to revise though. I was actually looking for stuff like this very hard on the internet, but I couldn’t obviously find them because well, they weren’t there yet
Thank you so much! I have my first group theory exam tomorrow and this is a huge help! I wish my Professor knew how to animate things hahaha!
Thanks Mathemaniac, this clarifies a lot.
One question: S3 (group of all symmetries of the triangle), how can we find its (normal) subgroup(s)? By Lagrange, the order of |H| \in {1, 2, 3, 6}, with {e} and {e, r, r2, f, rf, r2f} have orders 1 and 6. But for other subsets, e.g. {e, r, f}, why is this (not) a subgroup and why is (not) it normal?
My previous chapter about Lagrange's theorem said something about subgroups: the set has to be closed under composition, e.g. rf is not inside {e, r, f}, and so this is not a subgroup, let alone a normal one. {e, f} on the other hand is a subgroup, but not normal. Composing reflection with itself is the identity, and so this is a subgroup, but as discussed in the video, viewing f in a rotated perspective, i.e. rfr^(-1), is not e or f (but should be r^2 f), so this subgroup is not normal.
@@mathemaniac Is {e, r^2f} is a subgroup? Composing r^2f with itself gives r^2f*r^2f, but I don't see how this is either e or r^2f...
If I would check if it's normal, I'd for example do r r^2f r^(-1)?
You might have to learn more about how these elements are related to each other, specifically rs = sr^(-1). That might help.
The other way would be to recognize that f, rf, r^2 f are reflections along different axes. So a composition of two reflections would be e, and so {e, r^2 f} is a subgroup, but not normal. Yes, you can definitely check the subgroup is not normal by actually doing r (r^2 f) r^(-1), but the point of this video is that you can see it directly by thinking of these as symmetries.
@@mathemaniac Thanks. Yes, rs = sr^(-1) is true.
if I compose f and rf , the this is not equal to e. If I call the vertices in a triangle {A, B, C} and I first reflect vertex A followed vertex B, then I get {A, C, B} followed by {B, C, A}. What am I missing? Or should I somehow shift the perspective like in 3:35?
@Mathemaniac Would {e, s, r^(-1)}={e, s, s r^2} be normal? Because r s r^(-1) = r^(-1).
Why do you pronounce it like that :(
Hard to follow
You should read up on the subject and then watch the video.