If $X$ and $Y$ be graphs, then their product $X\times Y$ has vertex set $V(X)\times V(Y)$, and $(x,y)\sim (x', y')$ iff $x\sim x'$ and $y\sim y'$.

Now, Show that if $X\times X\simeq Y\times Y$, then $X\simeq Y$.

Two graphs $X$ and $Y$ are isomorphic if there is a bijection, $\varphi$ say, from $V(X)$ to $V(Y)$ such that $x\sim y$ in $X$ iff $\varphi(x)\sim \varphi(y)$ in $Y$.


  • $\begingroup$ Are your graphs finite? $\endgroup$ – Hagen von Eitzen Sep 20 at 12:03
  • $\begingroup$ No, for any graph $\endgroup$ – Farshad Hasani Sep 20 at 12:11
  • 3
    $\begingroup$ We must restrict to finite graphs because there are counterexamples in the infinite case: Let $a<0$ and $V(X)=\{\,x\in\Bbb Z\mid x>a\,\}$ and edges $x\sim y\iff x,y>0$. Then $X\times X$ consists of countably infinitely many isolated vertices, namely those $(x_1,x_2)$ with $x_1\le 0\lor x_2\le 0$ and the subgraph formed by all other vertices (i.e., with $x_1,x_2>0$) does not depend on $a$. Hence $X\times X$ is the same no matter which $a<0$ we started with. But the number of isolated vertices in $X$ itself does vary with $a$. $\endgroup$ – Hagen von Eitzen Sep 20 at 16:30
  • $\begingroup$ Your counterexample is good. Thank you. So how about for finite graphs? $\endgroup$ – Farshad Hasani Sep 20 at 17:34

For the finite case this is a result of Lovasz. Two facts, first if F is a fixed graph then the number of homomorphisms from $F$ to $X\times X$ is the square of the number of homomorphisms from $F$ to $X$. Second if the number of homomorphisms from $F$ to $X$ equals the number from $F$ to $Y$, then $X$ and $Y$ are isomorphic.


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