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A group of order thirty-six with a normal $C_3^2$ and a cyclic quotient of order four acting freely on its eight nonzero elements has six conjugacy classes of sizes $1,4,4,9,9,9$. Its four linear characters are lifted from the cyclic quotient. The remaining irreducible degrees are four and four, by the sum of squared degrees. Column orthogonality forces the last three values of both degree-four rows to vanish. The square of the order-four action is inversion on $C_3^2$, making the first two nontrivial class values real. Orthogonality then gives sum minus one and product minus two, so the rows are $(4,1,-2,0,0,0)$ and $(4,-2,1,0,0,0)$. This illustrates why reality must be justified when reconstructing a character table from orthogonality.

For the nontrivial action of $C_3$ on $C_7$, the conjugacy class sizes are $1,7,7,3,3$. Three linear characters factor through the quotient $C_3$, and the remaining two irreducible characters have degree three. Put $\omega =e^{2\pi i/3}$ and let $\alpha ,\beta =(-1\pm i\sqrt{7})/2$. With suitable class labels the table isOne can derive the last two rows by inducing nontrivial characters of $C_7$ along the two orbits of the order-three automorphism on their exponents. Their degree-three orbit sums vanish off $C_7$. On a generator they give the two sums of seventh roots over quadratic residues and nonresidues. Their sum is $-1$ and their product is $2$: the nine pairwise exponent sums between the two triples contain zero three times and every nonzero residue once. Thus they are the roots of $X^2+X+2$. Alternatively column orthogonality forces the zeros on the centralizers of order three, and orthogonality with the identity column interchanges $\alpha$ and $\beta$ in the last row. This finite table illustrates how centralizer sizes and a small number of given values determine missing character values.

The class sizes sum to $|G|=36$. A degree-one character is a homomorphism into ${\mathbb{C}}^{\times }$. The supplied ${\chi }_2$ has order four and ${\chi }_2^2={\chi }_1$, so $1,{\chi }_2,{\chi }_1,{\bar{\chi }}_2$ are four distinct irreducible characters. A finite group has as many complex irreducible characters as conjugacy classes, so there are exactly two more. Their degrees $d,e$ satisfy $d^2+e^2=36-4=32$, whose only positive-integer solution is $d=e=4$.

The column orthogonality give ${\sum }_{\chi }|\chi (g){|}^2=|C_G(g)|$. Each of the last three classes has centralizer order $36/9=4$, already exhausted by the four degree-one rows. Thus both new rows vanish on $C_4,C_5,C_6$.

To justify that their remaining values are real, put $P=\ker {\chi }_2$. The supplied values show $P=C_1\cup C_2\cup C_3$, a normal subgroup of order nine, with $G/P\cong C_4$. A group of order nine is Abelian, by the standard p-group center argument and the fact that a cyclic quotient by the center forces the group to be Abelian. If $P$ were cyclic of order nine, conjugation through $G/P$ would act by an automorphism of order dividing both four and $|\mathrm{Aut}(C_9)|=6$, hence dividing two. Its nontrivial conjugacy classes could not have size four. Therefore $P\cong C_3^2$.

Let $h$ map to a generator of $G/P$, and let $A$ be its conjugation action on the two-dimensional vector space $P$ over ${\mathbb{F}}_3$. The two nonzero orbits have length four, so $A^2$ fixes no nonzero vector, while $A^4=I$. Since $t^2-1$ has distinct roots in characteristic three, the involution $A^2$ is diagonalizable; the absence of a $+1$ eigenvector gives $A^2=-I$. Thus conjugation by $h^2$ inverts every element of $P$. Every element in $C_2,C_3$ is therefore conjugate to its inverse, so every character value there is real.

Write one new row as $(4,a,b,0,0,0)$. Its orthogonality with the trivial row gives $4+4a+4b=0$, so $a+b=-1$. Its squared character inner product one gives $16+4a^2+4b^2=36$, so $a^2+b^2=5$ and $ab=-2$. Thus $a,b$ are $1,-2$ in some order. Column orthogonality with the identity column forces the other row's entries to be interchanged. Up to relabelling these two new characters, the completed character table isThe reality argument prevents an unjustified assumption when solving the orthogonality equations. This is character table from a cyclic quotient and two elementary-abelian orbits.

The class equation implies that a nontrivial finite $p$-group has nontrivial center of a group. Here a center of order $9$ would make $G/Z(G)$ cyclic; a group with cyclic quotient by its center of a group is abelian, contradicting the hypothesis. ThusThe quotient of order nine must be noncyclic for the same reason. For $x\notin Z(G)$, $⟨x,Z(G)⟩$ has order at least nine and lies in the proper centralizer $C_G(x)$. Lagrange's theorem therefore forces $\boxed{|C_G(x)|=9}$. Also $1\ne G^{\prime }\le Z(G)$, so $G^{\prime }=Z(G)$ and all commutators are central of order dividing three.

Classification and splitting. First suppose $a$ has order nine. Then $N=⟨a⟩$ is normal of index three and $Z(G)=⟨a^3⟩$. For $b\notin N$, conjugation on $N$ is nontrivial, since otherwise $a,b$ would generate an abelian group. Its order divides three, so, after replacing $b$ by its inverse if needed, $ba{b}^{-1}=a^4$. Write $b^3=a^{3k}$. Central commutators and $3\mid \left(\genfrac{}{}{0ex}{}{3}{2}\right)$ give $(a^{i}b{)}^3=a^{3i}{b}^3$; choose $i=-k$ modulo three to obtain an element $c=a^{i}b$ of order three outside $N$. This provesThere is a unique resulting isomorphism type, since the two nontrivial actions of order three are interchanged by reversing the complement's generator.

If there is no element of order nine, every nonidentity element has order three. Choose $a,b$ whose images form a basis of $G/Z(G)$ and put $c=[a,b]$. These elements generate $G$, and $c$ generates $Z(G)$. The subgroup $N=⟨a,c⟩\cong C_3^2$ has index three and is normal. The line $⟨b⟩$ meets $N$ trivially and complements it. Conjugation by $b$ fixes $c$ and sends $a$ to $ac$ or $a{c}^{-1}$; choosing the central generator removes this sign. Thus there is one isomorphism type,with the nontrivial unipotent action on ${\mathbb{F}}_3^2$. These presentations are realized by the indicated semidirect products and have order 27. Their exponents are nine and three respectively, so they are not isomorphic. This proves both existence and completeness of the two cases.

Their ordinary character tables are the same, up to relabelling. In either group each of the 24 noncentral elements has conjugacy class of size three, giving eight noncentral classes and three central singleton classes. Each noncentral class is $xZ(G)$, since conjugates of $x$ differ from $x$ by central commutators and its class has size three. There are nine linear characters, arising from $G/G^{\prime }\cong C_3^2$, and eleven irreducible characters altogether. The two remaining degrees satisfy $d_1^2+d_2^2=27-9=18$, so both are three.

Put $Z(G)=⟨z⟩$ and $\omega =e^{2\pi i/3}$. Label the noncentral classes by nonzero $(i,j)\in {\mathbb{F}}_3^2$. For $(r,s)\in {\mathbb{F}}_3^2$ the nine linear characters have valuesAt a noncentral element, column orthogonality gives a squared column norm of nine. The nine linear characters already contribute nine, so both degree-three characters vanish there. By the Schur lemma, a central element has value $3{\omega }^k$ in an irreducible representation of degree three. The central character cannot be trivial, since that would factor the representation through the abelian quotient. Orthogonality of the identity and $z$ columns then forces the two nontrivial central characters to occur once each. The final two rows, on $1,z,z^2$ and the eight other columns, areThis description gives the same entire character table in both cases. The differing element orders are additional power-map data, which an ordinary character table need not determine.

The identity column gives the group order as the sum of squared irreducible character degrees:By column orthogonality, the squared norm of a column is the corresponding centralizer order. In source order these orders and the resulting conjugacy class sizes areFor example the fourth column has squared norm $1+1+(|\alpha {|}^2+|\beta {|}^2+|\gamma {|}^2)+1=9$, while the seventh has squared norm $1+1+(|\xi {|}^2+|\eta {|}^2+|\varphi {|}^2)=7$.

An element's order divides its centralizer order. Cauchy's theorem for finite groups provides an involution, and only $C_2$ has an even centralizer order. Hence that entire class consists of involutions. The last three classes consist of elements of order seven. In $C_3,C_4,C_5,C_6$ the possibilities are three and nine. The three real numbers $\alpha ,\beta ,\gamma$ are roots of $t^3-3t+1$, which has no rational root and is irreducible over $\mathbb{Q}$. A real character value at an element of order three belongs to $\mathbb{Q}(e^{2\pi i/3})\cap \mathbb{R}=\mathbb{Q}$. Therefore the irrational real entries in $C_4,C_5,C_6$ rule out order three there. Cubing any of these order-nine elements produces an element of order three, which must belong to the sole remaining class $C_3$. ThusThe three deductions below give nine Sylow subgroups for $2$ with pairwise trivial intersections. Their conjugation action is transitive. Fix one, $S$. If $T\ne S$, put $H=S\cap N_G(T)$. The product $HT$ is a $2$-subgroup, so maximality of $T$ gives $H\le T$. Trivial intersection then gives $H=1$. Thus the orbit of $T$ under $S$ has size eight, exhausting the other eight points. The point stabilizer $N_G(S)$ is therefore transitive on the remaining points: the action is doubly transitive.

For a doubly transitive action, its permutation character has the form $1+\theta$ with $\theta$ irreducible. Here $\theta (1)=8$, and the unique degree-eight row is ${\chi }_6$. Therefore the numbers of fixed points are the entries ofEvery nonidentity element consequently fixes at most two points. This also shows that the action is faithful.