# What're pairwise, vertex disjoint copies of $K_5$?

I need some help nailing some graph theory terminologies down. I know that a planar graph $$G$$, is a graph where no two edges intersect one another. A vertex disjoint graph $$G1$$ is a graph, where $$V(G1_1) \ \cap V(G1_2)=\emptyset$$, or each sub graph of $$G1$$ don't share any edges. $$K_5$$ is a non-planar graph which has $$5$$ vertices which're connected to one another by edges. However I'm unsure as to what pairwise, vertex disjoint copies of $$K_5$$ are. It would be great if someone can explain in intuitive English what that is.

Here is one $$K_5$$: a graph with $$5$$ vertices (which I've numbered $$1$$ to $$5$$), with an edge connecting each pair of vertices. Here are two vertex-disjoint copies of $$K_5$$. Ten vertices ($$1$$ to $$10$$). The first five form one copy of $$K_5$$, since each pair of them is connected by an edge. The second five form the other copy of $$K_5$$, again each pair connected by an edge. They are vertex-disjoint since no vertex in the first five is in the second five. BTW: your statement about a planar graph is faulty. A planar graph is one that can be drawn so that no two edges cross each other. This is a planar graph, even though there are crossing edges in the picture: because you can draw it so those edges don't cross: • So if you wanted to you could make as many vertex disjoint copies of $K_5$ as you wanted, as long as the vertices are different? Also what would pairwise vertex disjoint copies of $K_5$? – user686555 Jul 4 '19 at 10:28
• Pairwise vertex disjoint just means each pair of copies are vertex disjoint. – Robert Israel Jul 4 '19 at 11:56
• So pairwise vertex disjoint copies just refers to the graph having $x$ many pair copies of $K_5$ such that each copy is vertex disjoint from the others? In this case, if we say that there are $6$ pairwise vertex disjoint copies of $K_5$, then there are $6$ five form graphs and in total $30$ vertices. – user686555 Jul 4 '19 at 21:01
• Yes, that's right. – Robert Israel Jul 4 '19 at 22:55
• So what's the maximum amount of pairwise vertex disjoint copies of $K_5$ can $G$, $G$ is $5$ planar, have? I know on the minimum side, it's zero because if we have $K_5$ with a vertex $6$ connected to $5$ and a new vertex $7$, where $7$ is connected to $4$ and $5$. If you remove vertex $1$ to $5$ you're just left with vertex $6$ and $7$, which is not a copy of $K_5$. – user686555 Jul 4 '19 at 23:16