TITLE: How do we compute Aut(Z2 x Z2)?
QUESTION [8 upvotes]: I know that $Aut(Z_n)=u(n)$.
While computing $Aut(Z_4\times Z_2)=Z_2$ and $Aut(Z_2\times Z_2\times Z_2)=S_3$, I considered all possible cases and checked if it was a homomorphism. I am not sure if the above results are correct, I may have made a mistake.
Is there any shortcut or tips to compute Automorphism groups quicker? Any software that could compute Automorphism groups would be good too. Also, could experts please help to check the above two results as I am not sure if I have done them correctly.
Thanks!

REPLY [7 votes]: Note that $ℤ _ 2 \times ℤ_2=\langle(0,1),(1,0)\rangle=\langle(1,0),(1,1)\rangle=\langle(1,1),(0,1)\rangle$ and these are only only generators with minimum no. of elements(equivalent to basis of a vector space).
Now f is a automorphism on $ℤ_2 \times ℤ_2$ iff $f$ 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 $f({(0,1),(1,1)})={(0,1),(1,0)}$ . The rhs can be chosen 3 ways and each case can be done for 2 ways for $f((0,1))=(0,1)$ or $(1,0))$
So$ |\text{Aut}(ℤ_2 \times ℤ_2 )|=6$
Also note that the maximum order of a element this group is 3 i.e
$\text{Aut}(ℤ_2 \times ℤ_2) \cong S_3$