Determine whether this map $\phi$ is an isomorphism $\langle M_2 (\mathbb{R}), \cdot \rangle$ with $\langle \mathbb{R}, \cdot \rangle$ where $\phi(A)$ is the determinant of matrix $A$. 
I don't believe the map is an isomorphism because  I don't think it is injective (one-to-one). I know that if a matrix is not invertible, then the determinant will always be 0 and that the $\det(A) = \det(A^T)$. However I am having a hard time backing up my answer and explaining what I know to show that the map is not an isomorphism. Here is what I have so far.

Let $A$ be a $2 \times 2$ Matrix and let B be the transpose of A. Then $\phi(A) = \det(A) = \det(B) = \phi(B)$. 

Is this sufficient enough to show that the map is not injective? 
 A: Others have commented on the proof that you have presented. I will present an alternative proof in case it proves helpful.
Let us suppose that $\phi: M_2(\mathbb R) \to \mathbb R$ is an isomorphism. Then its inverse, $\phi^{-1}: \mathbb R \to M_2(\mathbb R)$ must also be an isomorphism. 
Now, pick two matrices $H_1, H_2 \in M_2(\mathbb R)$ such that $H_1 H_2 \neq H_2 H_1$. Since $\phi^{-1}$ is an isomorphism, it must also be surjective, so there exists elements $n_1, n_2 \in \mathbb R$ such that $\phi^{-1}(n_1) = H_1$ and $\phi^{-1}(n_2) = H_2$. Therefore, we have
$$
H_1 H_2 = \phi^{-1}(n_1) \phi^{-1}(n_2) = \phi^{-1}(n_1 n_2)
$$
Since real numbers commute (meaning that the group $(\mathbb R, \cdot)$ is abelian), $n_1 n_2 = n_2 n_1$. Hence,
$$
\phi^{-1}(n_1 n_2) = \phi^{-1}(n_2 n_1).
$$
Once more applying the properties of an isomorphism gives us
$$
\phi^{-1}(n_2 n_1) = \phi^{-1}(n_2)\phi^{-1}(n_1) = H_2 H_1.
$$
Therefore,
$$
H_1 H_2 = H_2 H_1.
$$
which contradicts the condition that $H_1 H_2 \neq H_2 H_1$, so our assumption that $\phi$ is an isomorphism must be false.
A: To show a map is not injective, all you need to do is find two inputs that map for the same output. In this case, the following matrices have the same determinant:
\begin{align*}
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
, &&
\begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix}
\end{align*}
So the map is not injective (hence not an isomorphism).
A: Suppose it's a group isomorphism. Then there is a group homomorphism $\psi$ s.t. $\psi (\phi(x)) = x$ and $x = \phi(\psi(x))$. Take any two real numbers, $a$ and $b$. Since $ab$ = $ba$, $\psi(ab) = \psi(ba)$. $\psi(ab) = \psi(a)\cdot\psi(b) = \psi(b)\cdot\psi(a)$. Since $a$ and $b$ are arbitrary and $\psi$ is bijection, $\psi(a)$ and $\psi(b)$ are arbitrary, thus for any matrices $A$ and $B$, they commute. This is false though.
