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Jan 24, 2015 at 2:34 comment added Eric Wofsey @TimothyChow: It is not hard to see that if $G\cong L(G)$ and $G$ has a vertex of degree at least 3, then $G$ has a vertex of degree at least $n$ for all $n\in\mathbb{N}$. Basically, start with a graph with a vertex of degree 3 and one more connected edge and then iterate the operation $L$; all the graphs obtained from this must embed in $G$, but they have vertices of arbitrarily high degree.
Jan 23, 2015 at 20:10 comment added Timothy Chow I'm wondering if maybe your real question is about graphs of bounded degree?
Jan 23, 2015 at 19:59 comment added Timothy Chow Regarding the side question, the complete graph with a countable number of vertices is isomorphic to its line graph.
Jan 23, 2015 at 9:24 vote accept Dominic van der Zypen
Jan 23, 2015 at 9:11 history edited Dominic van der Zypen CC BY-SA 3.0
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Jan 23, 2015 at 9:10 comment added Dominic van der Zypen That's right - > just edited
Jan 23, 2015 at 8:57 answer added Eric Wofsey timeline score: 6
Jan 23, 2015 at 7:05 history asked Dominic van der Zypen CC BY-SA 3.0