If k and n-k-1 are coprime then strongly regular graph G is imprimitive

Question

I am trying to solve the exercise 10.1 from Algebraic graph theory by Godsil and Royle. I’m stuck with it, can anyone give me some hint. Below is the question.

Suppose G is a strongly regular graph on $n$ vertices with degree of regularity $k$, then G is imprimitive if $n$ and $n-k-1$ are coprime.

Answer 1

$\textit{Proof.}$ Consider the following claim.

• $\textit{Claim:}$ The following are equivalent,

  1. $X = (n,k,a,c)$ is a strongly regular graph that is disconnected.
  2. $c=0$
  3. $k = a+1$
  4. All connected components of $X$ are isomorphic to $K_{k+1}$.

  $\textit{Proof:}$ As in a strongly regular graph, the number of vertices common to vertices $a$ and $b$ which are not connected is equal to $c$. Now if the graph is disconnected then vertices in two differnt components cannot be adjacent to any vertex and hence $c=0$. As $c=0$, any two neighbors of $v \in V$ should be adjacent, so $a = k-1 \implies k=a+1$. Hence all the vertices in a connected component should be adjacent. Therefore each connected component is $K_{k+1}$.

Now consider a vertex $v$ in the given strongly regular graph $G(n,k,a,c)$. If $c \neq 0$, each of the $k$ neighbours of $u$ is adjacent to $u$ itself, to a neighbours of $u$, and thus to $k - a - 1$ non-neighbours of $u$, for a total of $k(k - a -1)$ edges. On the other hand, each of the $n - k -1$ non-neighbours of $u$ is adjacent to $c$ neighbours of $u$ for a total of $(n- k- 1)c$ edges. Therefore, $k(k-a-1) = (n-k-1)c$.

Assuming $c \neq 0$ in $G$, else it is trivially imprimitive as per claim. Consider $k(k-a-1) = (n-k-1)c$. As all $n, k, a$ and $c$ are integers, $k \ \lvert \ (n-k-1)\cdot c$. But $gcd(k, n-k-1) = 1$. Hence $k \ \lvert \ c$. But in any strongly regular graph, $k \geq c$. As $c \neq 0$, we should have $k = c$. But this implies $\overline{G}(n,n-k-1,n-2-2k+c,n-2k+a)$ is disconnected as $\overline{k} = n-k-1 = n-2k+k-1 = n-2k+c-1 = n-2k-2+c + 1 = \overline{a} + 1$, by the claim, we have $\overline{G}$ is disconnected. So, if the g.c.d. of $k$ and $n-1-k$ is 1, then $G$ is imprimitive.

If $G$ is imprimitive and $G$ is disconnected, $G$ is a disjoint union $\lambda K_m$ of $\lambda$ complete graphs of size $m$ (and $v = \lambda m$, $k = m - 1$, $a = m - 2$, $c = 0$), as explained in the claim. If $\overline{G}$ is disconnected, $\overline{G}$ is a disjoint union of complete graphs which implies $G$ is a complete multipartite graph $K_{\lambda \times m}$ (and $v = \lambda m$, $k = (\lambda - 1)m$, $a = (\lambda - 2)m$, $c = (\lambda - 1)m$).