Commutative graph product with multiplicative independence number? Given two graphs $G,H$ is there a product $\star$ such that 1. and 2. holds where $\alpha$ refers to independence number? 


*

*$$\alpha(G\star H)=\alpha(G)\alpha(H)=\alpha(H\star G)$$

*$$G\star H\cong H\star G$$
Lexicographic product satisfies 1. but not 2.
Also other than Lexicographic product is there any other product that satisfies 1.?
 A: The "disjunctive" or "OR" product works (see https://en.wikipedia.org/wiki/Graph_product). For graphs $G$ and $H$, define $G * H$ as the graph with vertex set $V(G) \times V(H)$, with two vertices $(g,h)$ and $(g',h')$ being adjacent if $g \sim g'$ or $h \sim h'$, where $\sim$ denotes adjacency. Obviously $G * H \cong H * G$. It is also easy to see that if $S_G$ and $S_H$ are independent sets in $G$ and $H$, then $S_G \times S_H$ is an independent set in $G * H$, and thus $\alpha(G * H) \ge \alpha(G)\alpha(H)$. Conversely, if $S$ is an independent set in $G * H$, then define the sets
\begin{align*}
S_G &= \{g : \exists h \in V(H) \text{ s.t. } (g,h) \in S\}; \\
S_H &= \{h : \exists g \in V(G) \text{ s.t. } (g,h) \in S\}.
\end{align*}
These are just the "projections" of $S$ onto $G$ and $H$ respectively. By the definition of $G * H$ and the fact that $S$ is an independent set, we have that $S_G$ and $S_H$ are independent sets in $G$ and $H$ respectively. Also, $S \subseteq S_G \times S_H$, and thus $\alpha(G * H) \le \alpha(G)\alpha(H)$ as desired.
Note that this is related to the so-called "strong" product, denoted by $\boxtimes$, by the equation $G * H = \overline{\overline{G} \boxtimes \overline{H}}$. The lexicographic product is a subgraph of the disjunctive product.
