Conjugations of a group by an element x can be seen as the image of an inner automorphism Cx on the group. Since the image of the 4-group under Cx must be the 4-group, as DrBob said, the 4-group must be closed under conjugations hence it is normal.
@@MathDoctorBob I knew the physics' definition and you said rotations only in video too. What is bothering me that, D_8 has a subgroup of rotations and the elements are e,(1234),(13)(24),(1432). Aren't they the only rotations for a square? Since we can make finite rotations by 90° only in D_8 ?
@@MathDoctorBob yeah , but my point is these are they only 4 rotations I can think of. How the other rotations came there ? Don't we have only four rotations while each rotations can be 90° only?
Hi! @ 9:10 I think you made a mistake in listing the elements of A4. |A4|=12 not 3. Also that invalidates coset cardinality since all cosets must contain same number of elements.
at 1:00, you state that inner automorphisms map the subgroup to itself, so therefore, the subgroup is normal. I understand that INN(G) is a normal subgroup of AUT(G), but why does that mean the subgroup of A4 is normal to A4? wouldn't your reasoning mean every subgroup of any group with an inner automorphism is normal to the group? thank you again, Dr. Bob
Sir I have one question.... Automorphism of Dn is isomorphic to which group ?? (Here Dn is dihedral group). I have only idea of its order.... Order of Aut(Dn) is n × phi(n)
Your welcome! INN(G) normal in AUT(G) is too much for here. The 4-group H is closed under conjugations by G; this just means ghg-1 is in H for any h in H and g in G, or just the definition of normal. It is also normal because it is the only subgroup of order 4.
Sir is this not a method to calculate # of automorphisms : "Sending generators of a group to a generator of the same group"? And now here you are sending generators of A_4 to same ordered elements and claiming that the # of automorphism is less than the product of choices.So how is this possible? Is this bcoz the sent elements or the images may not produce the whole group? So is there no way to find all the sets of generators of any general group as then I think that will be the equal to |Aut(G)|
It's not enough to send a generating set to another generating set - the relations between them need to hold also. Homomorphism is part of automorphism. What happens here is that the maximum number of possible automorphisms by checking orders of elements accounts for everything before considering relations.
@@MathDoctorBob ok so if the relations too hold then sending gens to gens will be Aut(G).Now if we send only same ordered element generator to same ordered element generator,without caring about the relations between them(as you have shown in video),then will it be bijection or homomorphism?
In this case, the argument is special; we know conjugation by elements of S_4 gives 24 elements (once we check no conjugation fixes every element of A_4.) The count without homomorphism, just on orders of elements, is at most 24. So we have the order without needing to check the homomorphism property. In general, you do need to check, which makes finding Aut(G) a difficult problem.
![[Deprecated] Group Theory Lecture 5.5 Automorphisms of Permutation Groups](/img/1.gif)
Conjugations of a group by an element x can be seen as the image of an inner automorphism Cx on the group. Since the image of the 4-group under Cx must be the 4-group, as DrBob said, the 4-group must be closed under conjugations hence it is normal.
What does rigid motion means?
Preserves distances and orientation. In this case, rotations only.
@@MathDoctorBob
I knew the physics' definition and you said rotations only in video too. What is bothering me that, D_8 has a subgroup of rotations and the elements are e,(1234),(13)(24),(1432). Aren't they the only rotations for a square? Since we can make finite rotations by 90° only in D_8 ?
@@sudiptoborun D_8 contains 4 reflections also.
@@MathDoctorBob yeah , but my point is these are they only 4 rotations I can think of. How the other rotations came there ? Don't we have only four rotations while each rotations can be 90° only?
@@sudiptoborun D_8 is not the same as A_4.
Hi! @ 9:10 I think you made a mistake in listing the elements of A4. |A4|=12 not 3. Also that invalidates coset cardinality since all cosets must contain same number of elements.
That's a shorthand indicating the cycle types. So (ab)(cd) = (12)(34),(13)(24),(14)(23).
MathDoctorBob Oh cool! Sorry, my bad.
at 1:00, you state that inner automorphisms map the subgroup to itself, so therefore, the subgroup is normal. I understand that INN(G) is a normal subgroup of AUT(G), but why does that mean the subgroup of A4 is normal to A4? wouldn't your reasoning mean every subgroup of any group with an inner automorphism is normal to the group? thank you again, Dr. Bob
Your reasoning is correct provided that there are no other subgroups of the same order, but not otherwise.
Sir I have one question....
Automorphism of Dn is isomorphic to which group ??
(Here Dn is dihedral group).
I have only idea of its order.... Order of Aut(Dn) is n × phi(n)
Your welcome! INN(G) normal in AUT(G) is too much for here. The 4-group H is closed under conjugations by G; this just means ghg-1 is in H for any h in H and g in G, or just the definition of normal. It is also normal because it is the only subgroup of order 4.
What assumptions do we have? In general, not true; see Z/5.
My professor asked us this in class: If G has no element of order three, then G has a subgroup of order 4, why?
Thanks
Sir is this not a method to calculate # of automorphisms :
"Sending generators of a group to a generator of the same group"?
And now here you are sending generators of A_4 to same ordered elements and claiming that the # of automorphism is less than the product of choices.So how is this possible? Is this bcoz the sent elements or the images may not produce the whole group?
So is there no way to find all the sets of generators of any general group as then I think that will be the equal to |Aut(G)|
It's not enough to send a generating set to another generating set - the relations between them need to hold also. Homomorphism is part of automorphism. What happens here is that the maximum number of possible automorphisms by checking orders of elements accounts for everything before considering relations.
@@MathDoctorBob ok so if the relations too hold then sending gens to gens will be Aut(G).Now if we send only same ordered element generator to same ordered element generator,without caring about the relations between them(as you have shown in video),then will it be bijection or homomorphism?
In this case, the argument is special; we know conjugation by elements of S_4 gives 24 elements (once we check no conjugation fixes every element of A_4.) The count without homomorphism, just on orders of elements, is at most 24. So we have the order without needing to check the homomorphism property. In general, you do need to check, which makes finding Aut(G) a difficult problem.