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The matrix-block sizes in the Artin–Wedderburn theorem for $\mathbb{C}[G]$ recover the irreducible character degrees with their multiplicities. They determine the degree-square sum and the number $|G:G^{\prime }|$ of linear characters. Isaacs' character-degree criterion for p-nilpotency therefore detects each normal p-complement and applying it for every prime also detects nilpotent groups. The set of distinct degrees without multiplicities does not contain the same information.

For a finite group $G$ and prime number $p$, the notationdenotes the irreducible characters whose irreducible character degree is coprime to $p$.

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.

The implication is false. ChooseThe hook lengths areThus $3$ does not divide this irreducible character degree, but no hook has length $a{p}^k=6$.

There is no contradiction with part (i): removing two horizontal rim hooks of length three gives the empty core of a partition, so $w_3(\lambda )=2=a$ and the core degree is one. Having total weight of a partition two does not force a single hook of length twice the modulus.