Let $ A $ be a finite subgroup of $ SU_{n_1} $ such that the inclusion $ A \hookrightarrow SU_{n_1} $ is an irrep. And let $ B $ be a finite subgroup of $ SU_{n_2} $ such that the inclusion $ b \hookrightarrow SU_{n_2} $ is an irrep. Then $$ A \otimes B = \{ a \otimes b: a \in A,b \in B \} $$ is a subgroup of $ SU_{n_1n_2} $. Is the inclusion $ A \otimes B \hookrightarrow SU_{n_1n_2} $ an irrep? Also what can we say about the surjection $$ A \times B \to A \otimes B $$ is the kernel just all the $ e^{i\theta} $ such that both $ e^{i \theta} I_{n_1} \in A $ and $ e^{i \theta} I_{n_2} \in B $?
Let $ \chi_A $ be the character for the inclusion of $ A $ and let $ \chi_B $ be the character for the inclusion of $ B $. Since both are irreps we have $$ 1= \frac{1}{|A|}\sum_a |\chi_A(a)|^2 $$ and $$ 1= \frac{1}{|B|}\sum_b |\chi_B(b)|^2 $$ Let $ \chi $ be the character of the inclusion of $ A \otimes B $. Then the inclusion is an irrep if and only if $ <\chi,\chi>=1 $. I would like to say something like $$ <\chi,\chi> = \frac{1}{|A \otimes B|} \sum_{a \otimes b} \chi(a \otimes b) \\ = \frac{1}{|A \otimes B|} \sum_{a \otimes b} \chi_A(a) \chi_B(b) \\ = \frac{1}{|A \times B|} \sum_{(a,b)} \chi_A(a) \chi_B(b) \\ = \big( \frac{1}{|A|}\sum_a |\chi_A(a)|^2 \big) \big( \frac{1}{|A|}\sum_b |\chi_B(b)|^2 \big) \\ =1 $$
However the third line is kind of fishy, it sort of gets back to my question about the relationship between $ A \otimes B $ and $ A \times B $.
In your character calculation you can do everything for the group $G =A\times B$ to show that it’s an irreducible rep of $G$. (Which certainly implies it’s an irred of the quotient of $G$ which acts faithfully.
You are also right about the kernel of the map from $G$ to the tensor product. You can just see this on matrices and the definition of how the tensor product of two matrices acts on the tensor product of the vector spaces. In particular, if $v$ is not an eigenvector of $b$ then $w \otimes v$ is not an eigenvector of $a \otimes b$ so that is non-trivial. hence the only way it can be trivial is if every vector is an eigenvector of $b$ (similarly $a$) which means they must both be scalars whose product is $1$.