Let $G$ be a group, and $X$ a group homomorphism from $G$ to $GL_d(\mathbb{C})$. Define a function
$\phi:G/N → GL_d(\mathbb{C})$ by $\phi(gN) = X(g)$
where $N = {\{g \in G: X(g) = I}\}$.
show that $\phi$ is a well defined function and that $\phi$ is faithful.
attempt: Let $x = g_1N$ and $y = g_2N$, then $x = y$ implies $g_1N = g_2N$, then $\phi(g_1N) = \phi(g_2N)$ implies $X(g_1) = X(g_2)$ implies $I = I$ So $\phi$ is well defined.
And recall that $\phi$ is faithful, if it's injective. And $\phi$ is injective if and only if the kernel $\phi$ is the identity subgroup of $G$.
So for all $\phi(x) =\phi(y)$ then $\phi(g_1N) = \phi(g_2N) $ implies $X(g_1) = X(g_2)$ implies $I = I$. So $\phi$ is injective.
Can someone please verify this, or help me if it's incorrect. Any feedback would really help . Thank you!
Here is a correct proof. I am relying on the fact that in any group $G$, two cosets $g_1H$ and $g_2H$ of a given subgroup $H$ are equal if and only if $g_1^{-1}g_2 \in H$.
Well defined: suppose $x = g_1N$ and $y = g_2N$. To say that $\phi$ is well defined is to say that if $x = y$, then $X(g_1) = X(g_2)$. Since $g_1N = g_2N$, you must have $g_1^{-1}g_2 \in N$. Hence $X(g_1^{-1}g_2) = I$. But then $$I = X(g_1^{-1}g_2) = X(g_1^{-1}) \cdot X(g_2) = X(g_1)^{-1} X(g_2)$$ so $X(g_1) = X(g_2)$.
Injectivity: the same argument backwards. Suppose $\phi(g_1N) = \phi(g_2N)$. We need to show that $g_1N = g_2N$. By hypothesis, $X(g_1) = X(g_2)$, so $X(g_1^{-1}g_2) = I$, hence $g_1^{-1}g_2 \in N$, hence $g_1N = g_2N$.