$\operatorname{Aut}(\mathbb{Z}_2 \times \mathbb{Z}_2) \cong \mathbb{Z}_2$

Question

Calculate $\operatorname{Aut}(\mathbb{Z}_2 \times \mathbb{Z}_2)$.

My solution: $ \operatorname{Aut}(\mathbb{Z}_2 \times \mathbb{Z}_2) \cong \operatorname{Aut}(\mathbb{Z}_4) \cong U(4) \cong \mathbb{Z}_2 $

Is my solution correct ?

Reasoning used: Any subgroup of order $4$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. So $\mathbb{Z}_2 \times \mathbb{Z}_2$ can be replaced by $\mathbb{Z}_4$. That $\operatorname{Aut}(\mathbb{Z}_4) \cong U(4)$ is a theorem, and $U(4) \cong \mathbb{Z}/2$.

Doubt: There are $4$ elements in $\mathbb{Z}_2 \times \mathbb{Z}_2$. The identity will go to the identity. From the remaining three, $6$ different functions can be formed. Therefore $$ \operatorname{Aut}(\mathbb{Z}_2 \times \mathbb{Z}_2) \cong S_3. $$ How is this possible?

Answer 1

What is wrong, is that every subgroup of order $4$ is not isomorphic to $\mathbb Z_4$. In particular, $A = \mathbb Z_2 \times \mathbb Z_2$ is not isomorphic to $\mathbb Z_4$ since it contains an element of order two, while $\mathbb Z_4$ does not, and every isomorphism takes an element to another element of the same order.

How you would argue this is as follows. Note that $\mathbb Z_2 \times \mathbb Z_2$ has four elements, namely $(0,0),(0,1),(1,0),(1,1)$ (where $0,1$ are the elements of $\mathbb Z_2$ i.e. congruence classes modulo two). We call these elements $e,a,b,ab$ respectively to shorten notation. Note that $a^2 = b^2 = (ab)^2 = e$, so every non-trivial element has order two. Furthermore, the group is abelian.

Any automorphism takes $e \to e$, so there we have no freedom.

Note that it is enough to decide where any automorphism takes $a$ and $b$, since $\phi(ab) = \phi(a)\phi(b)$.

So we have the following problem : assign to $a,b$, two numbers in such a manner that the above relation of $\phi$ is satisfied.

The claim is that this will be done all the time. That is, if you send $a,b$ to any two different elements out of $ab,a,b$, then $ab$ going to the third element is a valid automorphism.

Indeed, this is clear, since if you take any two elements of the set $\{ab,a,b\}$ and multiply them, you get the third element. For example, $ab \times a = b$. Thus, the size of $\mbox{Aut }(A)$ is six.

So let us give a map $\psi : S_3 \to \mbox{Aut }(A)$, as follows : Form a mini-map l : $\{1,2,3\} \to \{a,b,ab\}$ given by $l(1) = a,l(2) = b,l(3) = ab$ (like a label map), and $(\psi(\phi))(x) = l(\phi(x))$.

$\psi$ is well defined, since $\phi$ and $l$ are both bijective(on their respective domains/codomains), so $\psi$ is sending an automorphism to a bijection.

$\psi$ is a homomorphism : this is easy to see.

It is injective, since if $l(\phi(x)) = x$ for every $x$ then in particular $\phi(a) = a$ and $\phi(b) = b$ so $\phi$ is the identity map. Of course, since the sets are of the same size, this is a bijection.

Hence the automorphism group of the Klein-4 group is automorphic to the symmetric group on three elements.

Answer 2

$\mathbb{Z}^2 \times \mathbb{Z}^2$ is a non-cyclic group of order four.

so all non-identity elements are of order 2. let the non-identity elements be $a, b, c$. each non-identity element is its own inverse $a^2 = b^2 = c^2 =1$.

so $ab = c = ba$ because you can rule out $ab = a, ab = b$ and $ab = 1$ by what has already been stated, and likewise for $ba$

similarly $bc = a = cb$ and $ca = b = ac$

the complete symmetry wrt $a, b$ and $c$ means that these elements are interchangeable. i.e. the equations: $$ a^2 = b^2 = c^2 = 1 \\ ab = c = ba \\ bc = a = cb \\ ca = b = ac $$

are simply permuted by any permutation of $a, b$ and $c$.

Answer 3

As stated in the comments $\mathbb Z_2 \times \mathbb Z_2$ cannot "be replaced" by $\mathbb Z_4$ as they're not isomorphic.

Hint: You can think of $\mathbb Z_2 \times \mathbb Z_2$ as consisting of the elements: $\{(0,0);(1,0);(0,1);(1,1)\}$ with the group product (or sum, since it's abelian and it's usual to use "$+$") defined coordinate-wise. A group automorphism of $\mathbb Z_2 \times \mathbb Z_2$ must map the neutral element to itself i.e $(0,0) \mapsto (0,0)$. Now, for the remaining elements of the group, to which element can they map to?