I came across this question on brilliant. I first observed that any particular automorphism (i.e., a bijective homomorphism) must map each group element to a distinct element in the group set: if it were to map two elements to the same one, the inverse of the automorphism wouldn't exist. Therefore we can create a candidate automorphism by permuting the group elements (every permutation is invertible).
Then, we can find the group operation of the automorphism $f$ as follows: $\forall x, y \in f(G): x \cdot y = f(f^{-1}(x)) \cdot f(f^{-1}(y)) = f(f^{-1}(x) * f^{-1}(y))$. (The inverse exists because it is bijective and the last equality is the definition of homomorphisms.)
Subsequently I verified that $f(G)$ obeys associativity, identity and invertibility, and thus is indeed a group. In fact, this holds for every permutation (if my python code is correct).
Lastly, I observed that no two permutations lead to the same automorphism, therefore there is no double counting.
I thus came to the conclusion that there are $|Z_2 \times Z_2 \times Z_2|! = 8! = 40320$ automorphisms.
According to the brilliant though, the correct answer is 168. The justification there is:
An automorphism is determined uniquely by what elements the three
generators $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ turn into under the map.
The first one can turn into any of seven nontrivial elements. The
second one can turn into any of six elements (not the identity and not
what $(1,0,0)$ turned into). The third one can turn into any of four
elements (it can't turn into any of the four elements generated by the
images of the first two generators). So, the answer is $7 \cdot 6 \cdot 4 = 168$.In fact, the automorphism group is $GL_3({\mathbb F}_2)$, the group of
with coefficients in $\{0, 1\}$ modulo $2$.
Where did I go wrong?
Best Answer
You’ve made a subtle but very understandable mistake. You found the number of bijections from $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ where you could put the group structure of $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ on the image; however automorphisms don’t come with group structures, you need to work with the already existing one. For example, your definition would consider the map $f$ that swaps $(0, 0, 0)$ and ($1, 0, 0)$ and fixes everything else to be a valid automorphism. However, that map isn’t, since $f((0, 0, 0) + (0, 1, 0)) \neq f(0, 0, 0) + f(0, 1, 0)$.
The + on both sides needs to be doing the same thing for this to be an automorphism! More generally, for any homomorphism $g : G \to H$, you need $G$ and $H$ to already have group structured and for $\phi$ to respect the existing structure, not define a new one.