Consider the following subgroup of the group of matrices $(M_2(\mathbb{C}), +)$:
$$ G = \Bigg\{ \begin{pmatrix} z_1 & z_2\\ -\overline{z_2} & \overline{z_1}\\ \end{pmatrix} \Bigg | \hspace{.1cm} z_1, z_2 \in \mathbb{C} \Bigg\} \subseteq M_2(\mathbb{C})$$
I am asked to find an injective morphism between the groups $(\mathbb{C}, +)$ and $(G, +)$.
So what I understood is that I have to find a function $f$
$$f: \mathbb{C} \rightarrow G$$
that has the following properties:
$I. f(z_1 + z_2) = f(z_1) + f(z_2), \hspace{.3cm} \forall \hspace{.1cm} z_1, z_2 \in \mathbb{C}$
$II. f$ is injective
So then, without any strategy in particular, I started coming up with different functions and checked to see if they had the above properties. I came up with this function:
$$f(z) = \begin{pmatrix} z & z\\ 0 & 0\\ \end{pmatrix}$$
for $z \in \mathbb{C}$. I observed that:
$$f(z_1 + z_2) = \begin{pmatrix} z_1 + z_2 & z_1 + z_2\\ 0 & 0\\ \end{pmatrix}$$
$$f(z_1) + f(z_2) = \begin{pmatrix} z_1 & z_1\\ 0 & 0\\ \end{pmatrix} + \begin{pmatrix} z_2 & z_2\\ 0 & 0\\ \end{pmatrix} = \begin{pmatrix} z_1 + z_2 & z_1 + z_2\\ 0 & 0\\ \end{pmatrix} $$
So we have that $f(z_1 + z_2) = f(z_1) + f(z_2)$, therefore the function $f$ is a morphism and also the first property is satisfied. Now we have to check the second property.
If we would have $z_1, z_2 \in \mathbb{C}$ such that $f(z_1) = f(z_2)$, we would have:
$$\begin{pmatrix} z_1 & z_1\\ 0 & 0\\ \end{pmatrix} = \begin{pmatrix} z_2 & z_2\\ 0 & 0\\ \end{pmatrix}$$
And that would result in the fact $z_1 = z_2$ so the function is indeed injective. Since the statement of the problem is set up in such a way, I assumed that the function needs to be injective, while also not being surjective, so I also checked to make sure that $f$ is not surjective.
Say we have $A \in G$ such that $f(z) = A$. We would have:
$$\begin{pmatrix} z & z\\ 0 & 0\\ \end{pmatrix} = \begin{pmatrix} z_1 & z_2\\ -\overline{z_2} & \overline{z_1}\\ \end{pmatrix}$$
Which would only happen when $z_1 = z_2 = z = 0$ so I concluded that the function $f$ is not surjective.
So, given everything I have shown, we have that the function:
$$f: \mathbb{C} \rightarrow G \hspace{2cm} f(z) = \begin{pmatrix} z & z\\ 0 & 0\\ \end{pmatrix}$$
is an injective (and also not surjective) morphism between the groups $(\mathbb{C}, +)$ and $(G, +)$.
Is my work correct? I am full of doubt about it. It seems so random, I didn't even follow any approach to finding the function, I just did trial and error. And is this function that I found really an injective morphism? In a nutshell, is my solution correct?
$G$ contains the matrix, $J_z=\begin{pmatrix}z&0\\0&\bar z\end{pmatrix}$ , for any$z\in\Bbb C$.
Map $z\to J_z$.