Let $G,H$ be two groups and $\rho_G,\rho_H$ be irreducible representations of $G$ and $H$ on complex vector spaces $V_G,V_H$, respectively. Then in several sources, for example, in http://mathworld.wolfram.com/ExternalTensorProduct.html , it is claimed that the external tensor product $\rho_G\otimes \rho_H$ is an irreducible $G\times H$-representation. I don't understand why this should be true in general. What confuses me is that the definition of the external tensor product $$ (\rho_G\otimes \rho_H)(g,h)(v\otimes w)=(\rho_G(g)v)\otimes (\rho_H(h)w),\qquad (v,w)\in V_G\times V_H, $$ which is extended linearly to all of $V_G\otimes V_H$, obviously leaves the subspace of pure tensors invariant, in other words, the image of the canonical map $\otimes: V_G\times V_H\to V_G\otimes V_H$ is a $G\times H$-invariant subspace. However, unless either $V_G$ or $V_H$ are at most of dimension $1$, $\otimes$ is not surjective, so $\rho_G\otimes \rho_H$ is reducible.
Where am I making a mistake?
The image of $\otimes$ is not a subspace! Remember, $\otimes$ is bilinear, not linear. So its image is not closed under sums. In general, if $v,v'\in V$ and $w,w'\in W$, then $v\otimes w+v'\otimes w'$ cannot necessarily be written in the form $v''\otimes w''$.