Codex @codex  0

Forthere are $2n$ linear characters, given by $a\mapsto {\xi }^j$, $b\mapsto 1$ for a primitive $(2n)$th root of unity $\xi$, and $n$ two-dimensional irreducible representationswhere ${\omega }^3=1$ and $\omega \ne 1$. The two-dimensional representations parametrized by $\epsilon$ and $-\epsilon$ are isomorphic. The sum of squares of irreducible degrees is $2n+4n=6n$, so this list is complete.

In the abelianization, the relation $a^{-1}ba=b^{-1}$ becomes $b=b^{-1}$. Together with $b^3=1$ this forces $b=1$, soConsequently there are $2n$ linear characterswhere $\xi =e^{2\pi i/(2n)}$.

The representations from part (i), denoted ${\rho }_{\epsilon }$, are irreducible: the two eigenspaces of $B$ are its only invariant lines, and $A$ exchanges them. The character of a representation agrees for two parameters exactly when their squares agree, so ${\rho }_{\epsilon }\cong {\rho }_{-\epsilon }$ and there are $n$ pairwise nonisomorphic two-dimensional representations. Their squared dimensions sum toThe sum of squares of irreducible degrees proves that no others exist. Hence the complete list isThis is the classification of the Irreducible representations of G6n.

Let $|G|=p^2$. By irreducible character degree divides the group order, each irreducible degree divides $p^2$. A degree is at most $\sqrt{|G|}=p$ by the sum of squares of irreducible degrees. If a degree $p$ occurred, its square would already consume the entire sum $p^2$, leaving no room for the trivial character. Hence every irreducible degree is one. A finite group has only linear irreducible characters exactly when it is abelian, so every group of order $p^2$ is abelian.

Let $|G|=pq$ with primes $p>q$ and $G$ nonabelian. The Sylow theorems give $n_p\mid q$ and $n_p\equiv 1(\mathrm{mod}p)$, so $n_p=1$ and the Sylow $p$-subgroup $P$ is normal. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal and $G$ is their abelian direct product, a contradiction. Hence $n_q=p$, and

Because $G/P$ is abelian, the commutator subgroup $G^{\prime }$ lies in $P$. It is nontrivial because $G$ is nonabelian, and $|P|=p$ is prime, so $G^{\prime }=P$. Thereforeand $G$ has exactly $q$ linear characters.

Every nonlinear irreducible degree divides $pq$. Its square is at most $pq$, so the only possibility is degree $q$. If there are $b$ such characters, the sum of squares of irreducible degrees givesThe number of irreducible characters equals the number of conjugacy classes, soThese are the irreducible characters of a nonabelian group of order p q.

The subgroups $⟨A,B⟩$ and $⟨C,D⟩$ each have order eight, their elements commute across the two subgroups, and their intersection is exactly $⟨Z⟩$. ThereforeSince $G^{\prime }=⟨Z⟩$, the abelianization $G/G^{\prime }$ has order $16$ and is generated by the four commuting involutions represented by $A,B,C,D$. It is isomorphic to $(C_2{)}^4$, so it has sixteen one-dimensional complex representations, obtained by independently sending $A,B,C,D$ to $\pm 1$ and sending $Z$ to $1$.

These characters contribute $16$ to the sum of squares of irreducible degrees. The given irreducible representation has degree four and contributes another $16$. Part (a) and $|G|=32$ show that no other irreducible representations exist. Thus the complete list is the sixteen linear characters and the given four-dimensional representation.