# Show $\phi$ is well defined and injective function.

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!

• Why does $g_1N = g_2N$ imply $X(g_1) = X(g_2)$? You need to argue this in order to show that $\phi$ is well defined.
– D_S
Feb 29 '16 at 21:05
• Also, to show injectivity, you need to show that if $\phi(g_1N) = \phi(g_2N)$, then $g_1N = g_2N$. You don't appear to have shown this.
– D_S
Feb 29 '16 at 21:05
• For injectivity, I know $\phi(g_1N) = \phi(g_2N)$ implies $X(g_1) = X(g_2)$ implies $I = I$ , so from $I = I$ I would have to conclude $I = I$ has to implie $g_1N = g_2N$? Feb 29 '16 at 21:14
• You seem to have ommited the definition of $\;X\;$ . Is that some kind of representation of $\;G\;$ ? Also, shouldn't the definition of $\;N\;$ be $\;N=\{g\in G\;:\;X(g)=I\}\;?$ Feb 29 '16 at 21:52
• Yes, you were right! N is the kernel Feb 29 '16 at 23:31

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 $g_1N = g_2N$.