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I am looking for a reference in which I can find a proof of the following result.

A strongly regular graph is disconnected if and only if it is a disjoint union of complete graphs $K_n$ of the same size.

A strongly regular graph connected if and only if it is a distance-regular graph of diameter 2.

mathma
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1 Answers1

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A graph is $(k,\lambda,\mu)$-strongly regular if each vertex has degree $k$, every two adjacent vertices have $\lambda$ common neighbors, and every two non-adjacent vertices have $\mu$ common neighbors.

When $\mu = 0$, two non-adjacent vertices have no common neighbors either; in other words, if there is a path of length $2$ between two vertices, then they are adjacent. By induction we can show that for all $\ell>1$, if there is a path of length $\ell$ between two vertices, then they are adjacent. Therefore non-adjacent vertices must be in different connected components, which means every component is a clique.

When $\mu > 0$, the graph has diameter $2$ because any two non-adjacent vertices have at least one common neighbor. We can check that it's distance-regular by considering all possibilities.

  • If $d(v,w) = 1$ then there are $\lambda$ vertices at distance $1$ from $v$ and distance $1$ from $w$.
  • If $d(v,w) = 1$ then there are $k-1-\lambda$ vertices at distance $1$ from $v$ and distance $2$ from $w$ (these are the remainder of $v$'s $k$ neighbors, excluding $w$ and the $\lambda$ vertices at distance $1$ from $w$) and similarly $k-1-\lambda$ vertices at distance $2$ from $v$ and distance $1$ from $w$.
  • If $d(v,w) = 1$ then there are $n - \lambda - 2(k-\lambda-1) - 2 = n - 2k + \lambda$ vertices at distance $2$ from $v$ and distance $2$ from $w$: all vertices apart from the ones counted above (and $v$ and $w$ themselves).
  • If $d(v,w) = 2$ then there are $\mu$ vertices at distance $1$ from $v$ and distance $1$ from $w$.
  • If $d(v,w) = 2$ then there are $k-\mu$ vertices at distance $1$ from $v$ and distance $2$ from $w$ (these are the remainder of $v$'s $k$ neighbors, excluding the $\mu$ vertices at distance $1$ from $w$) and similarly $k-\mu$ vertices at distance $2$ from $v$ and distance $1$ from $w$.
  • If $d(v,w) = 2$ then there are $n - \mu - 2(k-\mu) - 2 = n - 2k + \mu - 2$ vertices at distance $2$ from $v$ and distance $2$ from $w$: all vertices apart from the ones counted above (and $v$ and $w$ themselves).
Misha Lavrov
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