Exercise 5.16: Overlap of two states in isospin representation
$$ \begin{eqnarray} \langle a\, b ; J \, M ; T \, M_T | a'\, b'; J'\, M'; T' \, M_T \rangle &=& \left[ \mathcal{N}_{ab}(JT) [c_a^\dagger c_b^\dagger]_{JM}^{T M_T} |{\rm CORE} \rangle \right]^\dagger \left[ \mathcal{N}_{ab}(J'T') [ c_{a'}^\dagger c_{b'}^\dagger]_{J' M'}^{T' M_T'} | {\rm CORE} \rangle \right] \\[12pt] &=&\mathcal{N}_{ab}(JT) \mathcal{N}_{ab}(J'T') \sum_{\substack{m_\alpha m_\beta m_\alpha' m_\beta' \\ m_{t\alpha} m_{t \beta} m_{t\alpha}' m_{t\beta} } } (j_a \, m_\alpha\, j_b \, m_\beta | J \, M ) \\[12pt] \nonumber & \times & (j_a' \, m_\alpha' \, j_b' \, m_\beta' | J' \, M' )( \frac{1}{2} \, m_{t\alpha} \, \frac{1}{2} \, m_{t\beta} | T\, M_T ) ( \frac{1}{2} \, m_{t\alpha}' \, \frac{1}{2} \, m_{t\beta}' | T'\, M_T' ) \\[12pt] \nonumber &\times & (\delta_{\alpha \alpha'} \delta_{\beta \beta'} - \delta_{\alpha \beta'} \delta_{\beta \alpha'}) \\[12pt] \nonumber &= & \mathcal{N}_{ab}(JT) \mathcal{N}_{ab}(J'T') \sum_{m_\alpha m_\beta m_{t\alpha} m_{t \beta} } \left[ (j_a^{} \, m_\alpha^{}\, j_b \, m_\beta | J \, M ) \right. \\[12pt] \nonumber & \times & (j_a \, m_\alpha \, j_b \, m_\beta | J' \, M' )( \frac{1}{2} \, m_{t\alpha} \, \frac{1}{2} \, m_{t\beta} | T\, M_T ) ( \frac{1}{2} \, m_{t\alpha} \, \frac{1}{2} \, m_{t\beta} | T'\, M_T' ) \\[12pt] \nonumber &\times & \delta_{a a'} \delta_{b b'} - (j_a \, m_\alpha\, j_b \, m_\beta | J \, M ) (j_b \, m_\beta \, j_a \, m_\alpha\, | J \, M ) \\[12pt] \nonumber &\times& \left. ( \frac{1}{2} \, m_{t\alpha} \, \frac{1}{2} \, m_{t\beta} | T\, M_T ) ( \frac{1}{2} \, m_{t\beta} \, \frac{1}{2} \, m_{t\alpha} | T'\, M_T' ) \delta_{a b'} \delta_{b a'} \right] &=& \mathcal{N}_{ab}(JT) \mathcal{N}_{ab}(J'T') \delta_{JJ'} \delta_{MM'} \delta_{TT'} \delta_{M_TM_T'} \left(\delta_{aa'} \delta_{bb'} - (-1)^{j_a + j_b - J} (-1)^{\frac{1}{2} + \frac{1}{2} - T} \delta_{ab'} \delta_{ba'} \right) \\[12pt] \nonumber &=& \mathcal{N}_{ab}(JT) \mathcal{N}_{ab}(J'T') \sum_{m_\alpha m_\beta m_{t\alpha} m_{t \beta} } (j_a^{} \, m_\alpha^{}\, j_b \, m_\beta | J \, M ) \\[12pt] \nonumber & \times & (j_a \, m_\alpha \, j_b \, m_\beta | J' \, M' )( \frac{1}{2} \, m_{t\alpha} \, \frac{1}{2} \, m_{t\beta} | T\, M_T ) ( \frac{1}{2} \, m_{t\alpha} \, \frac{1}{2} \, m_{t\beta} | T'\, M_T' ) \\[12pt] \nonumber &\times & \left( \delta_{a a'} \delta_{b b'} - (-1)^{j_a + j_b + J + \frac{1}{2} + \frac{1}{2} + T} \delta_{ab'} \delta_{ba'} \right) \\[12pt] \nonumber &=& ( \mathcal{N}_{ab}(JT) )^2 \delta_{JJ'} \delta_{MM'} \delta_{TT'} \delta_{M_TM_T'} \left( \delta_{aa'} \delta_{bb'} + (-1)^{j_a + j_b + J + T} \delta_{ab'} \delta_{ba'} \right), \end{eqnarray} $$
where $\delta_{\alpha \alpha'} \delta_{\beta \beta'} - \delta_{\alpha \beta'} \delta_{\beta \alpha'}$ is derived from $\langle {\rm CORE} | c_\beta c_\alpha c_{\alpha'}^\dagger c_{\beta'}^\dagger | {\rm CORE} \rangle$ and $(-1)^{2J} = (-1)^{2T} = 1$ is used in the last line because $J$ and $T$ are integer.
Exercise 5.17: Normalization factor in isospin representation
The normalization factor is given by the condition of $ \langle a\, b ; J \, M ; T \, M_T | a\, b; J\, M; T \, M_T \rangle = 1$. By employing the result of Exercise 5.16,
$$ \begin{eqnarray} ( \mathcal{N}_{ab}(JT) )^2 \left( \delta_{aa} \delta_{bb} + (-1)^{j_a + j_b + J + T} \delta_{ab} \delta_{ba} \right) = 1 \end{eqnarray}. $$
Then, $\mathcal{N}_{ab}(JT)$ is
\begin{eqnarray} \mathcal{N}_{ab}(JT) = \sqrt{\frac{1}{1 + (-1)^{j_a + j_b + J +T} \delta_{ab} } }. \end{eqnarray}
Here, the denominator cannot be 0. Hence for the case where $a=b$, we have $(-1)^{j_a + j_a + J +T} = (-1)^{2 j_a + J + T} = - (-1)^{J+T} \neq (-1)$. This implies that $(-1)^{J+T} = -1$, whose factor is only valid for $a=b$. Physically, for the case of $a=b$, this means that when $T=1$, only odd values of $J$ are allowed. On the other hand, when $T=0$, only even values of $J$ are possible.
Hence, the given $\mathcal{N}_{ab}(JT)$ can be rewritten as follows:
$$ \begin{eqnarray} \mathcal{N}_{ab}(JT) = \sqrt{ \frac{1}{ \delta_{aa} \delta_{bb} - (-1)^{J+T} \delta_{ab} } } = \frac{ \sqrt{ 1 - (-1)^{J+T} \delta_{ab} } }{ 1 - (-1)^{J+T} \delta_{ab} } = \frac{ \sqrt{ 1 - (-1)^{J+T} \delta_{ab} } }{ 1 + \delta_{ab} }. \end{eqnarray} $$
