Let us consider the group $SU(N)$. Recently watching a recorded lecture in group theory for physicists I've heard about the following. Take the definining representation of $SU(N)$, namely, $(\rho,V)$ where $V=\mathbb{C}^N$ and where $\rho : SU(N)\to \operatorname{GL}(\mathbb{C}^N)$ is just the inclusion map, i.e., for every $g\in SU(N)$ we have $\rho(g)$ act upon $v\in \mathbb{C}^N$ by usual matrix multiplication.
Now take the $r$-fold tensor product $V^{\otimes r}$. The representation $\rho$ naturally induces one representation on $V^{\otimes r}$ which is characterized on decomposable tensors as $$[\rho^{\otimes r}(g)](v_1\otimes\cdots \otimes v_r)=(\rho(g)v_1)\otimes\cdots \otimes (\rho(g)v_r).\tag{1}$$
On the other hand, the symmetric group $S_r$ naturally acts on $V^{\otimes r}$. Its action on decomposable tensors is $$\pi(\sigma)(v_1\otimes\cdots\otimes v_r) = v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(r)}.\tag{2}$$
We may further show that the two actions commute. This is clear on decomposable tensors. We just apply $\pi(\sigma)$ to (1) and apply $\rho^{\otimes r}(g)$ to (2) and compare. Since decomposable tensors span $V^{\otimes r}$ and since $\rho^{\otimes r}(g)$ and $\pi(\sigma)$ are linear, the two operations must commute applied to anything.
Now I've heard that because of this, if we decompose $V^{\otimes r}$ in a direct sum of irreducible representations of $S_r$ the irreducible factors also give rise to irreducible representations of $SU(N)$. In other words, decompose $$V^{\otimes r}=\bigoplus_\lambda {\cal W}_\lambda,$$
where ${\cal W}_\lambda$ are irreducible representations of $S_r$, then ${\cal W}_{\lambda}$ is invariant under $\rho^{\otimes r}$ and, moreover, the subrepresentation of $\rho^{\otimes r}$ so obtained is irreducible.
Why is that the case? Why the two actions commuting imply that the irreducible $S_r$ factors of $V^{\otimes r}$ with respect to the $\pi$ are also irreducible factors of $SU(N)$ with respect to $\rho^{\otimes r}$?
Can all irreducible representations of $SU(N)$ be found in this way? Namely, by taking the definining representation, taking one sufficiently big $r\in \mathbb{N}$ and decomposing $V^{\otimes r}$ into irreducible factors of the symmetric group?