\(U \oplus \langle 4 \rangle^{2} \oplus \langle -4 \rangle^{3}\)

Let \(\Lambda = \mathrm{II}_{3,19}\) be the K3 lattice and \(G \subset O(\Lambda)\) a finite group of symplectic isometries. Row 10 of Table 10.3 of Hashimoto gives the invariant lattice \(\Lambda^G\) for the group \(\mathfrak{G}_{10} \cong D_8\) of Table 10.2, and this record is \(\Lambda^G\).

Roots: write \(r = a e_{1} + c e_{2} + w\) with \(w\) in the span \(W\) of \(e_{3}, \dots, e_{7}\). \(b(e_{1}, e_{2}) = 1\) is odd, \(W\) is orthogonal to \(e_{1}\) and \(e_{2}\), and \(b(w, w) \in 4\mathbb{Z}\) for \(w\) in \(W\), because the Gram tensor of \(W\) has its diagonal components in \(4\mathbb{Z}\) and the others in \(2\mathbb{Z}\). So \(b(r, e_{2}) = a\), \(b(r, e_{1}) = c\) and \(b(r, r) = 2 a c + b(w, w)\). If \(a\) is odd and \(c\) is even, then \(b(r, r) \equiv 0 \bmod 4\), and \(b(r, r)\) divides \(2 b(r, e_{2}) = 2 a \equiv 2 \bmod 4\), which is impossible; the same holds with \(a\) and \(c\) exchanged. Each root is therefore in \(L' = \{a \equiv c \bmod 2\}\), which has index 2 in \(L\). \(e_{1} + e_{2}\) has \(b(r, r) = 2\) and \(b(r, L) = \mathbb{Z}\), and \(2\) divides \(2\). \(e_{1} - e_{2}\) has \(b(r, r) = -2\) and \(b(r, L) = \mathbb{Z}\), and \(-2\) divides \(2\). \(e_{3}\) and \(e_{4}\) have \(b(r, r) = 4\) and \(b(r, L) = 4\mathbb{Z}\), and \(4\) divides \(8\). \(e_{5}\), \(e_{6}\) and \(e_{7}\) have \(b(r, r) = -4\) and \(b(r, L) = 4\mathbb{Z}\), and \(-4\) divides \(8\). These roots are in \(L'\), and their invariant factors in \(L\) are \(1, \dots, 1, 2\), so they generate \(L'\): \(\mathbb{Z}\Phi(L) = L'\), of index 2 in \(L\).