Is there a graph product that is multiplicative in independence number? I know that Stable set cannot be approximated to constant factor. I saw a simple proof using OR product sometime back. I am unable to recall it. If anyone here knows what I am talking about could help fill my memory, that will be helpful.

Is there a graph product $\cdot$ such that $\alpha(G\cdot G)=\alpha(G)^2$ for all  graphs $G$ and $G\cdot G$ the product of $G$ with itself?

I only need an answer to inapproximability to constant factor and I believe there is a one step proof for this.

If we had such a product, then we have $\alpha(G\cdot G\cdot\cdots \cdot G\cdot G)=\alpha(G)^k$. If $\alpha(G\cdot G\cdot\cdots \cdot G\cdot G)$ can be approximated to constant factor $\sigma&gt;0$, then $\alpha(G\cdot G\cdot\cdots \cdot G\cdot G)=\sigma\hat\alpha(G\cdot G\cdot\cdots \cdot G\cdot G)=\alpha(G)^k$ and since $0&lt;\sqrt[k]{\sigma}&lt;\sigma$, we can get an approximation of $\alpha(G)$ that is better than $\sigma$ by $\sqrt[k]{\sigma\hat\alpha(G\cdot G\cdot\cdots \cdot G\cdot G)}=\alpha(G)$.
Consider $G = K_2$, for which $\alpha(G)=1$.  The only product defined in terms of subgraphs of $K_4$ and for which $\alpha(G \times G) = 1$, is the strong product (⊠).  This doesn't quite work, since $\alpha(C_5 ⊠ C_5) = 5$, which is not a perfect square.

However, denoting the lexicographic product by $\circ$, there is an easy proof due to Geller and Stahl that $\alpha(G \circ H) = \alpha(G)\alpha(H)$.

So you want the lexicographic product.