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 Maschke theorem says that for a finite group over a field whose characteristic does not divide its order, every finite-dimensional group representation is semisimple; equivalently its group algebra is a semisimple algebra. This applies over $\mathbb{C}$.

The Artin–Wedderburn theorem says that a finite-dimensional complex semisimple algebra is a product ${\prod }_i{M}_{d_i}(\mathbb{C})$, with one simple module ${\mathbb{C}}^{d_i}$ for each factor, up to isomorphism. Applying it to $\mathbb{C}G$ yieldsThe dimension of the left-hand side is $|G|$, while a matrix algebra $M_d(\mathbb{C})$ has dimension $d^2$. Consequently

For order $105=3\cdot 5\cdot 7$, Sylow's congruence and divisibility conditions giveIf neither Sylow subgroup were a normal subgroup, there would be $21\cdot 4=84$ elements of order five and $15\cdot 6=90$ elements of order seven. Distinct subgroups of either prime order intersect only at the identity, and elements of the two orders are disjoint. This would give at least $84+90+1=175$ elements in a group of order $105$, a contradiction. Thus a subgroup of order five or seven is a normal subgroup, and the group is not simple.

For the preliminary index assertion, the coset action gives a homomorphism $G\to S_n$. Its kernel is a normal subgroup and cannot be all of $G$, because the action is transitive on $n>1$ points. Simplicity makes the kernel trivial. The sign homomorphism of this faithful action must be trivial: otherwise simplicity would identify $G$ with $C_2$, impossible for a transitive action of degree at least three. Thus $G$ embeds in $A_n$, and Lagrange givesA group of order $112=16\cdot 7$ has a Sylow two-subgroup of index seven. If the group were simple, the preceding result would give $112\mid 7!/2=2520$. This fails because $112$ is divisible by $16$ while $2520$ has only three factors of two. Thus there is no simple group of order $112$.

For order $56$, Sylow gives $n_7\in \{1,8\}$. If it is one, the order-seven Sylow is a normal subgroup. Otherwise its eight Sylows of order seven intersect only at the identity, giving $8\cdot 6=48$ nonidentity elements of order seven. Only eight elements remain including the identity. Every Sylow two-subgroup has order eight and consists entirely of these remaining elements. Hence all Sylow two-subgroups coincide and the order-eight subgroup is a normal subgroup. Either case gives a nontrivial proper normal subgroup, so no group of order $56$ is simple.

The quotient $PN/N$ is a $p$-group, being the image of $P$, andIts index is prime to $p$, so $PN/N$ is Sylow in $G/N$.

If $P\le A$ and $Q\le B$ are Sylow $p$-subgroups, their product has the full $p$-part of $|A\times B|$, hence is Sylow. Conversely, for any Sylow $R\le A\times B$, its two projections are $p$-subgroups and lie in some Sylows $P,Q$. Then $R\subseteq P\times Q$, and equality follows by order. Distinct pairs give distinct subgroups because their projections recover $P$ and $Q$. Thus

A Sylow subgroup for the prime $p$ is a maximal $p$-subgroup; in a finite group of order $p^{a}m$ with $p\nmid m$ it has order $p^a$. Sylow theorems assert existence, containment of every $p$-subgroup in a Sylow $p$-subgroup, conjugacy of all Sylow $p$-subgroups, and the counting conditions $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

Since $N$ is a normal subgroup, $NP$ is a subgroup andThe final index is prime to $p$. Since $N\cap P$ is a $p$-subgroup, its index in $N$ being prime to $p$ means it contains the entire $p$-part of $|N|$. Hence $N\cap P$ is Sylow in $N$.

Complete reducibility and character orthogonality imply that the squared multiplicities sum to three. The trivial constituent has multiplicity one, so the remaining two units come from two distinct nontrivial constituents, each with multiplicity one. Their degrees add to $4-1=3$, so they are two and one. ThusThe two-dimensional constituent is the complexification of the geometric action on the plane. Its trace is $2$ on the identity, $-2$ on the half-turn, zero on the quarter-turns, and zero on every reflection. Its character norm is $(4+4)/8=1$, proving it irreducible. Alternatively, the span of the two coordinate functions on the vertex set is a two-dimensional invariant subspace of the permutation module with this geometric action. This identifies the requested representation rather than merely its degree.

The character of a finite-dimensional complex representation is ${\chi }_{\rho }(g)=\mathrm{Tr}\rho (g)$; it is irreducible if the representation is irreducible. The character inner product isDistinct irreducible characters are orthogonal and each has norm one. For a permutation representation, the diagonal entry of the permutation matrix at a basis vector is one exactly when that vector is fixed. Its trace therefore counts fixed points.

Let $r$ be rotation through $\pi /2$. Let $s_x,s_y$ be reflections in the coordinate axes, and $s_d,s_{d^{\prime }}$ reflections in the two diagonals. The eight values areOnly diagonal reflections fix vertices, two each. Hence

A noncommutative algebra may have a commutative quotient: take $A=K\times M_2(K)$, $B=K$, and $\phi (\lambda ,M)=\lambda$. This is a surjective unital homomorphism. Neither requested change of commutativity can occur for an isomorphism, since an isomorphism transfers the identity $ab=ba$ in both directions.

For the matrix classifications, write $E_{ij}$ for a matrix unit and $N=E_{12}+E_{23}$. Then $N^2=E_{13}$ and $N^3=0$. The complete square-zero sets and commutativity results areIndeed $A_1$ is diagonal. In $A_2$ and $A_5$ the two off-diagonal generators have all products zero, so multiplication has the form $(x,u)(x^{\prime },u^{\prime })=(x{x}^{\prime },x{u}^{\prime }+x^{\prime }u)$. The diagonal entries of a square force $x=0$, after which the off-diagonal part squares to zero. In $A_3=K[N]$, a square-zero element first has scalar coefficient zero and then $(yN+z{N}^2{)}^2=y^2{N}^2$, forcing $y=0$. This argument works also in characteristic two. In $A_4$, a square first forces both diagonal parameters to vanish, leaving precisely $K{E}_{12}$. Noncommutativity follows from $E_{22}{E}_{12}=0$ but $E_{12}{E}_{22}=E_{12}$.

The coordinate map $xI+y{E}_{12}+z{E}_{13}\mapsto xI+y{E}_{13}+z{E}_{23}$ is an isomorphism $A_2\to A_5$, by the common multiplication formula. Every other pair is nonisomorphic: $A_4$ is the only noncommutative member, and among the commutative members the dimensions of the square-zero sets are respectively zero, two, one, and two. These sets are vector spaces in these examples, and an algebra isomorphism must carry one onto the other. Thus the only distinct isomorphic pair is $A_2\cong A_5$.

An algebra homomorphism is a $K$-linear map preserving multiplication, and preserving the unit under the unital convention. An algebra isomorphism is a bijective algebra homomorphism; its inverse then preserves the same operations.

A commutative domain algebra may map into a noncommutative target: take $A=K$, $B=K\times M_2(K)$, and $\phi (\lambda )=(\lambda ,\lambda I_2)$. This is even an injective unital homomorphism. The matrix factor makes the target noncommutative over every field, since $E_{12}{E}_{21}\ne E_{21}{E}_{12}$.

In the cyclic quotient $R/(d)$, the kernel is formed by residue classes of polynomials $h$ such that $d\mid Xh$. If $X\nmid d$, coprimality forces $d\mid h$, so the kernel is zero. If $d=Xe$, divisibility is equivalent to $e\mid h$, and the residue classes of multiples of $e$ modulo $Xe$ form a one-dimensional $K$-space. HenceThis too is immediately readable in the cyclic case, even if zero is a repeated root. The zero-dimensional module has kernel dimension zero separately.

For $d(X)=X^n+a_{n-1}{X}^{n-1}+\cdots +a_0$, the determinant of its companion matrix is $(-1{)}^n{a}_0$. ThereforeThus invertibility is readable from the constant coefficient, with no factorization or elimination.