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This dicyclic group has classes $1$, $a^3$, $\{a,a^5\}$, $\{a^2,a^4\}$ and the two parity classes of $b{a}^j$, of sizes $1,1,2,2,3,3$. Its quotient by $⟨a^2⟩$ is cyclic of order four, supplying four linear characters with $a\mapsto (-1{)}^j$, $b\mapsto i^j$. Its two remaining irreducible representations have $a\mapsto \mathrm{diag}({\zeta }^k,{\zeta }^{-k})$ and $b\mapsto \left(\begin{array}{cc}0& (-1{)}^k\\ 1& 0\end{array}\right)$ for $k=1,2$, where $\zeta =e^{i\pi /3}$. The traces are $2\cos (km\pi /3)$ on $a^m$ and zero on $b{a}^m$. Distinct diagonal eigenlines are exchanged by $b$, proving irreducibility; the degree-square sum $4+4+4=12$ proves completeness.

Directly tracking the supplied permutations givesThus every group element reduces to $a^k$ or $b{a}^k$, $0\le k<6$. These are twelve distinct elements: $a$ has order six, while $b$ is not a power of $a$ because it exchanges the two six-element supports that every power of $a$ preserves. Therefore $|G|=12$. This is the dicyclic group with parameter three; the calculations give its order-twelve dicyclic character table.

Conjugation by $b$ inverts powers of $a$, while conjugation by $a$ changes the exponent in $b{a}^k$ by two modulo six. Conjugation by $b$ preserves that exponent's parity. The complete conjugacy classes areTheir sizes are $1,1,2,2,3,3$.

The normal subgroup $N=⟨a^2⟩$ has order three. The quotient is cyclic of order four, generated by $bN$, since $(bN{)}^2=aN$. Inflating its characters gives ${\chi }_j(b)=i^j$, ${\chi }_j(a)=(-1{)}^j$, for $j=0,1,2,3$. These are all the linear characters: in any one-dimensional representation, the relation $ba{b}^{-1}=a^{-1}$ forces the image of $a$ to have square one, so $a^2$ is in the kernel.

In the displayed class order, the full character table isTo construct the last two rows, put $\zeta =e^{i\pi /3}$ and, for $k=1,2$, setThe matrices obey all three relations. The distinct eigenlines of ${\rho }_k(a)$ are exchanged by ${\rho }_k(b)$, so the representation is irreducible. Its traces on $a^m$ are $2\cos (km\pi /3)$ and on $b{a}^m$ are zero, producing the rows above. Their weighted inner products are orthonormal, and the degree-square sum $4\cdot 1^2+2\cdot 2^2=12$ proves completeness.

From $ba{b}^{-1}=a^{-1}$ we obtain $ba=a^{-1}b$. Moving every occurrence of $b$ to the right and replacing $b^2$ by $a^n$ shows that every group element has the formThe element $b$ cannot lie in $⟨a⟩$: otherwise it would commute with $a$, forcing $a=a^{-1}$, contrary to $a$ having order $2n>2$. Thus the two cosets $⟨a⟩$ and $⟨a⟩b$ are disjoint and each has $2n$ elements. Hence every $n$-dicyclic group has order $4n$.

Existence is explicit. Put $\zeta =e^{\pi i/n}$ and takeThen $a$ has order $2n$,The $2n$ diagonal matrices $a^k$ and the $2n$ off-diagonal matrices $a^{k}b$ are distinct, so they form a dicyclic group of order $4n$.