The S_5 character table by fixed-tabloid reconstruction

The class sizes and $M$. By [F1], $z_{\rho }={\prod }_i{i}^{m_i(\rho )}{m}_i(\rho )!$ evaluates to $5,4,6,6,8,12,120$ on the classes $(5),(4,1),(3,2),(3,1,1),(2,2,1),(2,1,1,1),(1^5)$: for instance $z_{(3,1,1)}=3\cdot 1^2\cdot 2!=6$ and $z_{(2,2,1)}=2^2\cdot 2!\cdot 1=8$. Hence $\mathrm{\mid }{C}_{\rho }\mathrm{\mid }=120\mathrm{/}{z}_{\rho }$ is $24,30,20,20,15,10,1$ and $M=\mathrm{diag}(5,4,6,6,8,12,120)$.

The matrix $A$. By [F2] the cycle types in $S_{\lambda }$ are obtained by combining the cycle types of the blocks. The single-block counts used are: in $S_1$, type $1$: $1$; in $S_2$, types $1^2,2$: $1,1$; in $S_3$, types $1^3,2\cdot 1,3$: $1,3,2$; in $S_4$, types $1^4,2\cdot 1^2,3\cdot 1,2^2,4$: $1,6,8,3,6$; and in $S_5$ the counts are the class sizes $1,10,15,20,20,30,24$ of types $1^5,2\cdot 1^3,2^2\cdot 1,3\cdot 1^2,3\cdot 2,4\cdot 1,5$. Combining block counts for each $\lambda$ (for example, in $S_{(3,2)}=S_3\times S_2$ the identity gives one element of type $1^5$; a $3$-cycle together with the identity gives the $2$ elements of type $3\cdot 1^2$; a $3$-cycle with the transposition gives the $2$ elements of type $3\cdot 2$; a transposition in $S_3$ combined with the identity, or the identity in $S_3$ combined with the transposition of $S_2$, gives the $3+1=4$ elements of type $2\cdot 1^3$; and a transposition in $S_3$ combined with the transposition in $S_2$ gives the $3$ elements of type $2^2\cdot 1$) yields exactly the displayed matrix $A$.

The matrix $B$. By [F3] the entry $b_{\lambda \nu }$ is $\mathrm{\mid }{S}_{\nu }\mathrm{\mid }$ times the number of semistandard tableaux of shape $\lambda$ and content $\nu$; the diagonal is $b_{\nu \nu }=\mathrm{\mid }{S}_{\nu }\mathrm{\mid }={\prod }_i{\nu }_i!=(120,24,12,6,4,2,1)$, and the displayed nonzero entries are confirmed by the corresponding counts: $b_{(4,1),(3,1,1)}=6\cdot 2=12$, because a semistandard tableau of shape $(4,1)$ and content $(3,1,1)$ must place $2$ or $3$ in the single lower cell and then is determined; $b_{(3,2),(2,2,1)}=4\cdot 2=8$ and $b_{(2,2,1),(2,1,1,1)}=2\cdot 2=4$ are checked the same way. The whole displayed matrix $B$ satisfies the Gram equation with the displayed $A$ and $M$: each entry of $B^{\mathsf{T}}B$ is the finite sum ${\sum }_{\lambda }{b}_{\lambda \nu }{b}_{\lambda \mu }$, and comparing with ${\sum }_{\rho }{z}_{\rho }{a}_{\nu \rho }{a}_{\mu \rho }$ gives, for example, $(B^{\mathsf{T}}B{)}_{(1^5),(1^5)}=1^2+4^2+5^2+6^2+5^2+4^2+1^2=120=z_{(1^5)}=(AM{A}^{\mathsf{T}}{)}_{(1^5),(1^5)}$ and $(B^{\mathsf{T}}B{)}_{(4,1),(3,2)}=24\cdot 12+24\cdot 12=576=6\cdot 8\cdot 2+8\cdot 3\cdot 3+12\cdot 6\cdot 4+120=(AM{A}^{\mathsf{T}}{)}_{(4,1),(3,2)}$. By the uniqueness clause of [F3] the nonnegative upper-triangular matrix with this Gram matrix and positive diagonal is $B$, so no other matrix can occur.

The character table. Multiplying $B$ on the right by the inverse transpose of $A$ gives the displayed $C$; equivalently, $C{A}^{\mathsf{T}}=B$, and this is the identity ${\sum }_{\rho }{\chi }^{\lambda }(C_{\rho })a_{\nu \rho }=b_{\lambda \nu }$ of the theorem. Two full rows of the product check it: the row $(5)$ consists of ones, and ${\sum }_{\rho }{a}_{\nu \rho }=\mathrm{\mid }{S}_{\nu }\mathrm{\mid }=b_{(5),\nu }$ for every $\nu$; for the row $(3,2)=(0,-1,1,-1,1,1,5)$ one gets ${\sum }_{\rho }{\chi }^{(3,2)}(C_{\rho })a_{(3,2),\rho }=0\cdot 0+(-1)\cdot 0+1\cdot 2+(-1)\cdot 2+1\cdot 3+1\cdot 4+5\cdot 1=12=b_{(3,2),(3,2)}$ and ${\sum }_{\rho }{\chi }^{(3,2)}(C_{\rho })a_{(4,1),\rho }=(-1)\cdot 6+(-1)\cdot 8+1\cdot 3+1\cdot 6+5\cdot 1=0=b_{(3,2),(4,1)}$; the remaining rows are verified by the same finite sums from the displayed $A$ and $B$. By the theorem's determinant formula, $\det C={\prod }_{\lambda ⊢5}{\prod }_i{\lambda }_i=5\cdot 4\cdot 6\cdot 3\cdot 4\cdot 2\cdot 1=2880$, the product of all parts of all partitions of $5$.

The orthogonality spot check of the example statement: with the class sizes of step 1.1 and the rows of $C$, $⟨{\chi }^{(3,2)},{\chi }^{(2,2,1)}⟩=\frac{1}{120}(24\cdot 0\cdot 0+30\cdot (-1)\cdot 1+20\cdot 1\cdot (-1)+20\cdot (-1)\cdot (-1)+15\cdot 1\cdot 1+10\cdot 1\cdot (-1)+1\cdot 5\cdot 5)=\frac{1}{120}(0-30-20+20+15-10+25)=0$, and $⟨{\chi }^{(3,2)},{\chi }^{(3,2)}⟩=\frac{1}{120}(30\cdot 1+20\cdot 1+20\cdot 1+15\cdot 1+10\cdot 1+25\cdot 1)=1$; the entries are real, so the conjugate in [F4] plays no role. Thus the displayed table is orthonormal on the checked pairs, as a character table must be.