GT11.1. Automorphisms of A4

Abstract Algebra: We compute the automorphism group of A4, the alternating group on 4 letters. We have that Aut(G) = S4, the symmetric group on 4 letters, Inn(A4) = A4, and Out(A4)=Z/2. We note that the coset structure splits S4 into even and odd permutations.

[MarkLiuX]

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]

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.

[MathDoctorBob]

What assumptions do we have? In general, not true; see Z/5.

[boboorozco]

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

[The_Pi_Piper]

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)

[andrejnj]

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.

[marycavazos]

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