A difficulty in understanding a step in finding all irreducible representations of $S_{3}.$

Question

Here is a part of the proof:

But I do not understand why to find sub-representations it is enough subspaces of V invariant under the action of both $\sigma $ and $\tau$? ...... could anyone explain this for me please ? is it a theorem?

Answer 1

By definition, since $S_3$ is generated by $\sigma$ and $\tau$, we can write any element of $S_3$ as $\sigma^{a_1} \tau^{b_1} \cdots \sigma^{a_n} \tau^{b_n}$ for some integers $a_1, b_1, \ldots, a_n, b_n$. In fact, for $S_3$ we can write any element as $\sigma^a \tau^b$ for some $a, b, but that's not necessary for our purposes.

Now, suppose that $\sigma, \tau$ preserve $V$. Then, for any $v \in V$, $\tau \cdot v \in V$, and thus $\sigma \cdot (\tau \cdot v) \in V$. But by definition of a representation, this is just $(\sigma \tau) \cdot v$. But there's nothing special about the particular product $\sigma \tau$ here---the same argument applies just as well to any other product of $\sigma$ and $\tau$.