No, since it is generated by two elements, or not? @Krijn

$D_2=\langle a, s \mid a^2=1=s^2, asa=s^{-1}=s\rangle$

We have that $\mathbb Z_2 \times \mathbb Z_2=\{(0,0), (0,1), (1,0), (1,1)$ and $D_2=\{e, a, s, as\}$, right?

Could we define the following map:
$h:\mathbb Z_2 \times \mathbb Z_2\rightarrow D_2$ with
$(0,0) \mapsto e \\ (0,1) \mapsto a \\ (1,0) \mapsto s \\ (1,1) \mapsto as$
? @Krijn

Since each element of $\mathbb Z_2 \times \mathbb Z_2$ is mapped to one element of $D_2$, and since for each element of $D_2$ there is one element of $\mathbb Z_2 \times \mathbb Z_2$ that is mapped to that, we conclude that themapping is bijective, right? Or is the justification wrong?
We have that each element of order $4$ is abelian, right?
To show that the mapping is an homomorphism do we have to show that $h_1\circ h_2 (x)=(h_1\circ h_2)(x)$, or not? @Krijn

@robjohn So we have to show that $\widehat{x}=2 \pi i \xi$, right?

But how can we calculate the FT of x? I wanted to use the definition, but the result tends to $+\infty$. So we have to use districutions. Could you explain to me how?