Questions about definition of edges in affine building. I have a question about about edges in a building in the book of expander graphs by Alexander Lubotzky, page 69.
We know that if $L_1' \subseteq L_2'$ and $[L_2' : L_1'] = p$, then there is an edge connecting $[L_1']$ and $[L_2']$.
In the third line of page 69, it is said that $[L_2' : L_1'] = p$ implies $pL_2' \subseteq L_1'$ and $[L_1' : pL_2'] = p$. How could we prove this result? 
I think that $L_2' = \mathbb{Z}_p v_1 + \mathbb{Z}_p v_2$ for some basis ${v_1, v_2}$ of $\mathbb{Z}_p \times \mathbb{Z}_p$. $[L_2' : L_1'] = p$ implies that $L_2' = \cup_{x \in L_2'} xL_1'$ and the number of cosets is p. But I don't know how to prove that $pL_2' \subseteq L_1'$ and $[L_1' : pL_2'] = p$. Thank you very much.
 A: By definition, $\mathbb{Z}_p=\{\sum_{i\geq 0}c_i p^i: c_i \in \{0, \ldots, p-1\}\}$.
We have $L_2'=\{xv_1+yv_2:x, y \in \mathbb{Z}_p\}= \{(x, y):x, y \in \mathbb{Z}_p\}$. Define $S_i = \{(px + iy, y): x, y \in \mathbb{Z}_p\}$, $i=0, \ldots, p-1$, $S_p=\{(x, py): x, y \in \mathbb{Z}_p\}$. Then it is easy to verify that $S_i, i \in [0, p-1]=\{0, \ldots, p-1\}$, $S_p$ are closed under addition and multiplication by elements of $\mathbb{Z}_p$. Therefore they are sub-lattices of $L_2'$.  
Let $x_1, x_2 \in \mathbb{Z}_p$. Then $\{(x_1+px + iy, x_2+y): x, y \in \mathbb{Z}_p\}=\{(x_1 - ix_2+px + i(y+x_2), y+x_2): x, y \in \mathbb{Z}_p\} = \{(x'+px + iy, y): x, y \in \mathbb{Z}_p, x' \in [0, p-1]\}$. Therefore there are $p$ cosets of $S_i$ in $L_2'$. Hence the index of $S_i$ in $L_2'$ is $p$. Similarly, the index of $S_p$ in $L_2'$ is also $p$. $S_0, \ldots, S_p$ are the $p+1$ sublattices of $L_2'$ of index $p$.
Note that we also have sublattices $T_i = \{(x, py + ix): x, y \in \mathbb{Z}_p\}$, $i=0, \ldots, p-1$, of $L_2'$. But $T_i = S_j, i\in [0, p-1], ij=1$ (assume that $p$ is a prime. Therefore $j$ exists). Indeed, let $(x, py+ix) \in T_i$. Take $x' = -jy, y'=py+ix$. Then $px'+jy'=x$. Hence $(x, py+ix)=(px'+jy', y') \in S_j$. Therefore $T_i \subseteq S_j$. Similarly, $S_j \subseteq T_i$.
If $L_1'$ is a sublattice of $L_2'$ of index $p$, then $L_1'$ is some $S_i$, $i \in [o, p]$. $pL_2' = \{(px, py): x, y \in \mathbb{Z}_p\} = \{(p(x-iy)+ipy, py): x, y \in \mathbb{Z}_p\} \subseteq \{(px+iy, y): x, y \in \mathbb{Z}_p\} = S_i = L_1'$ (if $i\in [0,p-1]$. The case of $i=p$ is similar). We now compute the index of $pL_2'$ in $L_1'$. Suppose that $L_1'=S_i, i\in [0, p-1]$. Let $(px+iy, y) \in L_1'$. Then $(px+iy, y) + pL_2' = \{(px+iy+px', y+py'):x, y, x', y' \in\mathbb{Z}_p\}=\{(p(x+x'+iy_1)+iy_2, y_2+p(y'+y_1): x, y, x',y'\in \in\mathbb{Z}_p, y_1 \in [0, p-1], py_1+y_2=y\}=(iy_2, y_2) + pL_2'$. Therefore $pL_2'$ has $p$ cosets in $L_1'$.
