I am curious what proof method is most commonly used to show that a representation is faithful. I have found remarkably little online about this question..
It makes sense how to show that a representation is not faithful, you simply show that two elements map to the same element, thus preventing the representation from being injective.
The converse is not so easy, however. There seems not to be a such simple trick. It would make sense to work out every element and verify that it's mapped distinctly from all of the rest, but this isn't practical in larger groups.
One could show that kernel is trivial, but one would also have to show that each element that isn't the identity, doesn't map to the identity, so this also fails.
I am wondering: what is the standard method to prove that a representation is faithful?
Let us look at a simple example:
There is a presentation for the symmetric group $S_3$ given by $$S_3 = \left\langle a, b: a^2 = b^2 = 1,(ab)^3 = 1\right\rangle$$
with the following representation (at least I think it is faithful): $\theta: S_4 \to \text{GL}(2, \mathbb{C})$ satisfying
$$\theta(a) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\: \theta(b) = \begin{pmatrix} 1 & -1 \\ 0 & -1 \end{pmatrix}.$$
Here is the most elegant solution which I have found (for this particular example):
In principle, we need to show that $|\ker\theta| = 1$, but this is hard to check directly. Using the First Isomorphism Theorem, another way to do it is to show that $|\text{Im}\:\theta| = 6$.
In fact, we don't even need to do that. If we can show that $|\text{Im}\:\theta| > 3$, then the fact that $|\text{Im}\:\theta|$ divides $|G|$ shows that $|\text{Im}\:\theta| = 6$.
Thus, we need only find four different elements in the image.
This is easy enough for this example and really only consists of doing one computation:
\begin{align} \theta(1) &= \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\\ \theta(a) &= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ \theta(b) &= \begin{pmatrix} 1 & -1\\ 0 & -1 \end{pmatrix}\\ \theta(a)\theta(b) &= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & -1\\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & -1\\ 1 & -1 \end{pmatrix}. \end{align}
The other two elements (that we don't actually need to find) take just a little more work, if anyone's curious.
\begin{align} \theta(b)\theta(a) &= \begin{pmatrix} 1 & -1\\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 1\\ -1 & 0 \end{pmatrix}\\ \theta(a)\theta(b)\theta(a) &= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & -1\\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0\\ -1 & 1 \end{pmatrix}. \end{align}