TIFR-2012-Maths-B: 11

Question

True/False Question:

The automorphism group  $Aut\left ( \mathbb{Z}/2\times \mathbb{Z}/2 \right )$ is abelian.

Answer 1

https://math.stackexchange.com/questions/1452260/how-do-we-compute-autz2-x-z2/2367924

 

Note that Z2×Z2=⟨(0,1),(1,0)⟩=⟨(1,0),(1,1)⟩=⟨(1,1),(0,1)⟩ℤ2×ℤ2=⟨(0,1),(1,0)⟩=⟨(1,0),(1,1)⟩=⟨(1,1),(0,1)⟩ and these are only only generators with minimum no. of elements(equivalent to basis of a vector space). Now f is a automorphism on Z2×Z2ℤ2×ℤ2 iff ff sends any of these three basis-sets to another basis-sets.i.e. we have 3 choices of a basis-set to be image of f .Also for any of these choices we have 2 choices to pick a particular value of a element.i. e there are 6 automorphism(for example Consider f((0,1),(1,1))=(0,1),(1,0)f((0,1),(1,1))=(0,1),(1,0) . The rhs can be chosen 3 ways and each case can be done for 2 ways for f((0,1))=(0,1)f((0,1))=(0,1) or (1,0))(1,0)) So|Aut(Z2×Z2)|=6|Aut(ℤ2×ℤ2)|=6 Also note that the maximum order of a element this group is 3 i.e Aut(Z2×Z2)≅S3