15. Block Designs

Put

We can use the first description to find $|X|$: there are $b$ ways to choose $j\in B$, and then $|C_j|=k$ ways to choose $p\in C_j$, so $|X|=bk$. Alternatively, we can use the second description. There are $v$ ways to choose $p\in V$, and then $|R_p|=r$ ways to choose $j\in R_p$, so $|X|=vr$. By comparing these, we see that $bk=vr$. Now put

We can again use the first description to find $|Y|$: there are $b$ ways to choose $j\in B$, then $k$ ways to choose $p\in C_j$, then $k-1$ ways to choose a different element $q\in C_j$, giving $|Y|=bk(k-1)$. Alternatively, we can use the second description: there are $v$ ways to choose $p\in V$, then $v-1$ ways to choose a different element $q\in V$, then $|R_p\cap R_q|=\lambda$ ways to choose $j\in R_p\cap R_q$, giving $|Y|=\lambda v(v-1)$. By comparing these, we get $bk(k-1)=\lambda v(v-1)$. We can now substitute our first equation $bk=vr$ into our second equation $bk(k-1)=\lambda v(v-1)$ and then divide by $v$ to get $r(k-1)=\lambda (v-1)$. Rearranging this, we get $\lambda /r=(k-1)/(v-1)$. As one of our axioms we assumed that , so , so , so . ∎

It will be harmless to assume that $V=\{1,\mathrm{\dots },v\}$ and $B=\{1,\mathrm{\dots },b\}$.

For $p=1,\mathrm{\dots },v$ we let ${𝐫}_p$ be the $p$’th row of the incidence matrix, so ${𝐫}_p\in {ℝ}^b$ with

We also put $𝐞=(1,1,\mathrm{\dots },1)\in {ℝ}^b$. We claim that

Indeed, as $𝐞=(1,1,\mathrm{\dots },1)$, the dot product ${𝐫}_p\cdot 𝐞$ is just the sum of the entries in ${𝐫}_p$, which is $|R_p|=r$. Similarly, the $j$’th term in ${𝐫}_p\cdot {𝐫}_p$ is $1^2=1$ if $j\in R_p$ and $0^2=0$ if $j\notin R_p$, so ${𝐫}_p\cdot {𝐫}_p=|R_p|=r$ again. On the other hand, if $q\ne p$ then the $j$’th term in ${𝐫}_p\cdot {𝐫}_q$ is $1$ if $j\in R_p\cap R_q$ and zero otherwise, so ${𝐫}_p\cdot {𝐫}_q=|R_p\cap R_q|=\lambda$. If we multiply this relation by $r$ and multiply the relation ${𝐫}_p\cdot 𝐞=r$ by $\lambda$ and subtract, we get ${𝐫}_p\cdot (r{𝐫}_q-\lambda 𝐞)=0$ in the case $q\ne p$. A similar argument gives ${𝐫}_p\cdot (r{𝐫}_p-\lambda 𝐞)=r^2-\lambda r=r(r-\lambda )$, as claimed. Note also that Proposition 15.4 gives $r>\lambda >0$, so $r(r-\lambda )\ne 0$.

We next claim that the vectors ${𝐫}_1,\mathrm{\dots },{𝐫}_v\in {ℝ}^b$ are linearly independent. Indeed, suppose we have a linear relation

For any $q$, we can take the dot product with the vector $r{𝐫}_q-\lambda 𝐞$, giving ${\sum }_{p=1}^v{\alpha }_p{𝐫}_p\cdot (r{𝐫}_q-\lambda 𝐞)=0$. The dot product relations proved above show that all the terms on the left are zero apart from the term where $p=q$; we therefore get ${\alpha }_{q}r(r-\lambda )=0$. As $r(r-\lambda )\ne 0$ this gives ${\alpha }_q=0$. This works for all $q$, so ${\alpha }_1=\mathrm{\cdots }={\alpha }_v=0$. This proves linear independence.

It is a basic fact of linear algebra that the maximum possible length of a linearly independent list in ${ℝ}^b$ is the dimension $b$. Thus, we must have $v\le b$. ∎

Put $U=(\mathbb{Z}/p)\setminus \{0\}$ for brevity. We can define a homomorphism $\alpha :U\to U$ by $\alpha (u)=u^2$, and then $Q$ is the image of $\alpha$ (which is one way to see that it is a subgroup). The First Isomorphism Theorem shows that $Q\simeq U/\ker (\alpha )$ and so $|Q|=|U|/|\ker (\alpha )|$. Here

As $\mathbb{Z}/p$ is a field, the product of two terms can only be zero if one of the terms is zero, so the equation $(u-1)(u+1)=0$ can only hold if $u=\pm 1$. This shows that $\ker (\alpha )=\{1,-1\}$, so $|Q|=|U|/2=2n+1$. We next claim that $-1\notin Q$. Indeed, if we have $-1=u^2$ then $u$ would be an element of order $4$ in $U$, but that is impossible (by Lagrange’s Theorem) because the order $|U|=2n+2$ is not divisible by $4$. Next, if $-i$ and $i$ were both in $Q$ then the element $-1=-i.i^{-1}$ would also be in $Q$, which is false. Thus, each of the sets $\{1,-1\},\{2,-2\},\mathrm{\dots },\{2n+1,-(2n+1)\}$ contains at most one element of $Q$. As $|Q|=2n+1$, we see that each of these sets must contain precisely one element of $Q$. ∎

Recall from Lemma 15.12 that either $x$ or $-x$ is a square. Suppose for the moment that $x$ is a square. Suppose that $(u,v)\in D_1$, so $u$ and $v$ are squares with $u-v=1$. It is clear that the product of two squares is a square, so $ux$ and $vx$ are squares with $ux-vx=x$, so $(ux,vx)\in D_x$. Conversely, if $(u^{\prime },v^{\prime })\in D_x$ then $(u^{\prime }/x,v^{\prime }/x)\in D_1$. From this it is clear that $|D_x|=|D_1|$.

Now suppose instead that $-x$ is a square. If $(u,v)\in D_1$ then $-vx$ and $-ux$ are squares with $(-vx)-(-ux)=(u-v)x=x$, so $(-vx,-ux)\in D_x$. Conversely, if $(u^{\prime },v^{\prime })\in D_x$ then $(-v^{\prime }/x,-u^{\prime }/x)\in D_1$. From this it is again clear that $|D_x|=|D_1|$.

We now see that $|D_x|=|D_1|$ in all cases, and the number of possibilities for $x$ is $p-1=4n+2$. The equation $|D|={\sum }_x|D_x|$ now becomes $|D|=(4n+2)|D_1|$. However, we saw previously that $|D|=(4n+2)n$, so $|D_1|=n$, so $|D_x|=n$ for all $x$. ∎

All that is left is to show that $|R_l\cap R_m|=n$ for all $l\ne m$. Recall that $R_l=l-Q$, so $j\in R_l$ iff $l-j\in Q$. Thus, if $j\in R_l\cap R_m$ we see that $l-j,m-j\in Q$ and of course $(l-j)-(m-j)=l-m$ so $(l-j,m-j)\in D_{l-m}$. We can therefore define a map $f:R_l\cap R_m\to D_{l-m}$ by $f(j)=(l-j,m-j)$. In the opposite direction, suppose that $(u,v)\in D_{l-m}$, so $u,v\in Q$ with $u-v=l-m$ or equivalently $l-u=m-v$. If we put $j=l-u=m-v$ then we find that $j\in R_l$ (because $R_l=l-Q$ and $j=l-u$) and also $j\in R_m$ (because $R_m=m-Q$ and $j=m-v$), so $j\in R_l\cap R_m$. Using this we see that $f$ is a bijection, so $|R_l\cap R_m|=|D_{l-m}|$. We also know from Lemma 15.14 that $|D_{l-m}|=n$, so $|R_l\cap R_m|=n$ as required. ∎