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.


  • 1
    $\begingroup$ Do you have two or three copies of $Z_2$? (For two its is $S_3$ for three not.) Also the autogroup of $Z_4 \times Z_2$ is rather larger. There are various results on the automorphism groups of (commutative) groups, can you explain what you want more specifically? $\endgroup$
    – quid
    Sep 26, 2015 at 15:06
  • $\begingroup$ @quid Could you explain (briefly) how to compute the two Automorphism groups? I have three copies of $Z_2$, I must have made a mistake somewhere. $\endgroup$
    – yoyostein
    Sep 26, 2015 at 15:24
  • 2
    $\begingroup$ The automorphism group of $Z_2^3$ is just $GL_3(Z_2)$. So all invertible $3\times 3$ matrices with entries from the field with two elements. I do not know off-hand another description for that group. But it is certainly quite a bit larger. note that you cannot only permute the element of some basis. You can also map $e_1$ to $e_1+ e_2$ and keep the rest fixed etc. By contrast $GL_2(Z_2)$ happens to be isomorphic to $S_3$. $\endgroup$
    – quid
    Sep 26, 2015 at 15:38
  • 3
    $\begingroup$ Possible duplicate of Follow-up to question: Aut(G) for G = Klein 4-group is isomorphic to $S_3$ $\endgroup$ Nov 6, 2016 at 21:50
  • 2
    $\begingroup$ Also a duplicate of this MSE-question. A shortcut is to use GAP, if you want. $\endgroup$ Nov 6, 2016 at 21:53

1 Answer 1


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 $\text{Aut}(ℤ_2 \times ℤ_2) \cong S_3$

  • 1
    $\begingroup$ @SRJ Consider elements of the group $Z_2 \times Z_2$ , under the operation addition you will notice their order is 3 at max $\endgroup$
    – The Doctor
    Jan 21, 2019 at 11:36

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