a First show that D4 has a normal subgroup of order 2 in fac

a) First, show that D₄ has a normal subgroup of order 2 (in fact, it is a quite special type of normal subgroup...). Note that D_n for all n even has such a subgroup, and for n odd there is no such subgroup.

b) Show that the group of automorphisms of the group D₄, denoted by Aut(D₄), contains at least 4 elements

Solution

D4 has 8 elements:

1,r,r²,r³, d₁,d₂,b₁,b₂ where r is the rotation on 90⁰ , d_1,d₂ flips over diagonals and b_1,b₂ are flips about the line joining the centers of opposite sides of square. Let N be a normal subgroup of D4. Note that

d₁₌rd₂r^-1 , b₁₌rb₂r^-1 and d₁d₂=b₁b₂=r²

hence if d₁ belongs to N and d₂ also belongs to N and similalry b₁ and b_2. thus if N contains a flip and N\ eqG , then N either contains d₁, d₂ or contains b₁ or b₂. Let N contains d1, d2 then N={1,d1, d2 , r² } and if N contains b₁ and b₂ then N={1, b₁ , b₂ ,r² }. Finally if N doesn\'t contain flips then N={1,r,r²,r³} or N={1, r² } thus D4 has one 2- element normal subgroup and three 4-element subgroups.thus a always D4 has two normal subgroups {1} and D4