# Tensor Product of Two Group Representations (Action Well-definedness)

I am studying the basic of representation. Coming back from https://en.wikipedia.org/wiki/Tensor_product_of_representations, I am having trouble filling out details why tensor product of two (finite dimensional) representations $$V$$ and $$W$$ over $$\mathbb{C}$$ of a finite group $$G$$ is also a representation.

I know Is the tensor product of two representations a representation? has an answer but I cannot understand such "high level" reasoning yet.

From the first link above, the action is defined as $$g.(v\otimes w):=(g.v)\otimes (g.w)$$, but I am not sure about its well-definedness.

Question 1: This definition cannot be seen as $$g$$ acting on arbitrary element in the tensor product, right? I mean, isn't arbitrary element in the tensor product a linear combination? Hence, what would the "true" action of $$g$$ be?

I think if I assume the that $$g$$ acts on an arbitrary element by linearity (meaning it can pass through the sigma and scalars), then here is my attempt based on my understanding (but I am not sure maybe something is wrong):

We first choose a basis $$\{v_i\}$$ and $$\{w_j\}$$ for $$V$$ and $$W$$, respectively. Thus, all $$\text{dim } V\times \text{dim } W$$ elements $$v_i \otimes w_j$$ form a basis for $$V\otimes W$$. To show well-definedness, I take elements $$a,b\in G$$ and two more elements in $$V \otimes W$$, but the elements in this tensor product is not simply basic tensor, right? Hence, I take $$\sum c_{i,j}(v_i\otimes w_j),\sum d_{i,j}(v_i\otimes w_j)\in V\otimes W$$ instead. If $$a=b$$ and $$\sum c_{i,j}(v_i\otimes w_j)=\sum d_{i,j}(v_i\otimes w_j)$$, then by linear independence, $$c_{i,j}=d_{i,j}$$ for each $$i,j$$. Hence, $$\sum c_{i,j}((a.v_i)\otimes (a.w_j)) =\sum d_{i,j}((a.v_i)\otimes (a.w_j))\text{ (by c_{i,j}=d_{i,j})}\\=\sum d_{i,j}((b.v_i)\otimes (b.w_j))\text{ (by well-definedness of each G-action on V and W)}.$$

Thus, the action map is well-defined. I think showing it satisfies the action property + linearity should be OK.

Question 2: Is the above reasoning correct?

Any suggestion or guidance is really appreciated. Thanks!.

A representation $$V$$ of $$G$$ is a homomorphism $$\rho :G \to GL(V)$$, where $$GL(V)$$ is the group of automorphisms of V.
Then, if $$g(v \otimes w)=g(v)\otimes g(w)$$ gives a representation $$V\otimes W$$ of $$G$$, then it should be an automorphism of $$V\otimes W$$.
It's not necessary to check arbitrary elements of the tensor product. Indeed, the rule defines the action of $$g$$ only on elements in the form $$v\otimes w$$ because, although omitted from the definition sometimes, it's linear, so $$g(\sum_i \alpha_i (v_i\otimes w_i))=\sum_i \alpha_i g(v_i\otimes w_i).$$
Now, the action is also well-defined, because $$g((\alpha_1 v_1+\alpha_2v_2)\otimes w) = g(\alpha_1 v_1+\alpha_2 v_2)\otimes g(w) = (\alpha_1 g(v_1)+ \alpha_2 g(v_2))\otimes g(w)$$ $$= \alpha_1 g(v_1)\otimes g(w) +\alpha_2g(v_2)\otimes g(w) = \alpha_1g(v_1\otimes w) + \alpha_2 g(v_2\otimes w)$$ and in the same way, $$g(v\otimes (\alpha_1 w_1+ \alpha_2w_2)) = \alpha_1g(v\otimes w_1) +\alpha_2g(v\otimes w_2).$$
Also, $$ker(g)$$ is trivial: $$g(v\otimes w) = 0 \implies g(v)\otimes g(w)=0 \implies g(v)=0\ \lor \ g(w)=0$$ But since $$g$$ is an automorphism on $$V$$ and $$W$$, then $$v=0$$ or $$w=0$$ and $$v\otimes w=0$$.
Since $$g$$ is linear and $$ker(g)$$ is trivial, then $$g$$ is an automorphism of $$V\otimes W$$. The fact that $$g^{-1}$$ is inverse of $$g$$ as an automorphism follows from definition on $$V$$ and $$W$$ and can be checked directly, thus the action of $$G$$ is isomorphic to a subgroup of $$GL(V\otimes W)$$, and $$V\otimes W$$ is a representation of $$G$$ by $$g(v\otimes w) = g(v)\otimes g(w)$$.