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


1. $$\alpha(G\star H)=\alpha(G)\alpha(H)=\alpha(H\star G)$$
2. $$G\star H\cong H\star G$$


Lexicographic product satisfies 1. but not 2.


Also other than Lexicographic product is there any other product that satisfies 1.?



Commutative graph product with multiplicative independence number?
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.


