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This is an interesting question, even though it is not well defined. Call a group "good" if it has a "good" bijection between its conjugacy classes and its irreducible complex representations. I agree with Alexander that the definition of a "good" bijection/group should be guided by classes of examples, but I prefer that a class of bijections/groups should be infinite.

There are families of good metacyclic groups. For example, if $n=2m+1$ is an odd integer, then the dihedral groups $D_{2n}=\langle a,b\mid a^2=b^n=1,\; a^{-1}ba=b^{-1}\rangle$ of order $2n$ are good. The conjugacy classes $\{b^j,b^{-j}\}$, $1\leq j\leq m$, $\{1\}$, and $\{a,ab,ab^2,\dots,ab^{n-1}\}$ correspond bijectively (I believe this is "good") to the irreducible representations $\rho_j$, $1\leq j\leq m$, $\sigma_0$, and $\sigma_1$, respectively where $\rho_j(a)=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\rho_j(b)=\begin{pmatrix}\zeta_n^j&0\\0&\zeta_n^{-j}\end{pmatrix}$, $\sigma_k(a)=\begin{pmatrix}(-1)^k\end{pmatrix}$, $\sigma_k(b)=\begin{pmatrix}1\end{pmatrix}$ and $\zeta_n=e^{2\pi i/n}$.

If an infinite family $G_1, G_2,\dots$ of groups is good, then you know a vast amount about each $G_n$ and can likely produce is a formula writing $|G_n|$ as a sum of the squares of the degrees of the irreducible representations. For $D_{2n}$ this is $2n=4m+2=m\times 2^2+2\times 1^2$. If $G_n$ is an extraspecial group of (odd) order $p^{1+2n}$ and exponent $p$, then the formula is $p^{1+2n}=(p-1)\times(p^n)^2+p^{2n}\times 1^2$. Perhaps the existence of such a formula should be part of the elusive definition of "good".

Addition: Yes Alexander, you are correct, the extraspecial group $G_n$ of order $p^{1+2n}$ and odd exponent $p$ is "good". To describe a "good" bijection I need some notation. Let $f_n\colon V\times V\to\mathbb{F}_p$ be a nondegenerate symplectic form on $V=\mathbb{F}_p^{2n}$. Multiplication in $G_n=V\times \mathbb{F}_p$ is given by $(v_1,\lambda_1)(v_2,\lambda_2)=(v_1+v_2,\lambda_1+\lambda_2+{\frac12}f_n(v_1,v_2))$, or by the matrices you indicate. The conjugacy classes are as follows: the $p$ one-element (central) classes $Z_\lambda:=\{(0,\lambda)\}$, $\lambda\in\mathbb{F}_p$, and the $p^{2n}-1$ classes $C_v:=\{(v,\lambda)\mid \lambda\in\mathbb{F}_p\}$ where $0\neq v\in V$. The irreducible representations are also easy. The trivial representation corresponds to the trivial character (or degree-1 representation). The $p^{2n}-1$ nontrivial characters correspond to the $p$-element classes $C_v$. The remaining $p-1$ irreducibles of degree $p^n$ correspond to the $p-1$ central conjugacy classes $Z_\lambda$, with $0\neq\lambda\in\mathbb{F}_p$. Fix a maximal totally isotropic subspace $W$ of $V$. By Witt's theorem $|W|=p^n$. Then $A:=W\times\mathbb{F}_p$ is a maximal abelian subgroup of $G_n$ of index $p^n$. Let $\sigma_\lambda$ be the 1-dimensional representation of $A$, with kernel $W$, mapping $(0,1)$ to $e^{2\pi i\lambda/p}$. The induced representations $\rho_\lambda={\rm Ind}_A^{G_n}(\sigma_\lambda)$ are irreducible of degree $p^n$. (A direct calculation shows $\langle\rho_\lambda,\rho_\lambda\rangle=1$. Choosing a different maximal totally isotropic subspace $W'$ gives equivalent representations $\rho'_\lambda$. The $W$s are permuted by ${\rm Aut}(G_n)$.) This is a "good" bijection, as identifying $V$ with its dual seems allowed.

This is an interesting question, even though it is not well defined. Call a group "good" if it has a "good" bijection between its conjugacy classes and its irreducible complex representations. I agree with Alexander that the definition of a "good" bijection/group should be guided by classes of examples, but I prefer that a class of bijections/groups should be infinite.

There are families of good metacyclic groups. For example, if $n=2m+1$ is an odd integer, then the dihedral groups $D_{2n}=\langle a,b\mid a^2=b^n=1,\; a^{-1}ba=b^{-1}\rangle$ of order $2n$ are good. The conjugacy classes $\{b^j,b^{-j}\}$, $1\leq j\leq m$, $\{1\}$, and $\{a,ab,ab^2,\dots,ab^{n-1}\}$ correspond bijectively (I believe this is "good") to the irreducible representations $\rho_j$, $1\leq j\leq m$, $\sigma_0$, and $\sigma_1$, respectively where $\rho_j(a)=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\rho_j(b)=\begin{pmatrix}\zeta_n^j&0\\0&\zeta_n^{-j}\end{pmatrix}$, $\sigma_k(a)=\begin{pmatrix}(-1)^k\end{pmatrix}$, $\sigma_k(b)=\begin{pmatrix}1\end{pmatrix}$ and $\zeta_n=e^{2\pi i/n}$.

If an infinite family $G_1, G_2,\dots$ of groups is good, then you know a vast amount about each $G_n$ and can likely produce is a formula writing $|G_n|$ as a sum of the squares of the degrees of the irreducible representations. For $D_{2n}$ this is $2n=4m+2=m\times 2^2+2\times 1^2$. If $G_n$ is an extraspecial group of (odd) order $p^{1+2n}$ and exponent $p$, then the formula is $p^{1+2n}=(p-1)\times(p^n)^2+p^{2n}\times 1^2$. Perhaps the existence of such a formula should be part of the elusive definition of "good".

Addition: Yes Alexander, you are correct, the extraspecial groups $G_n$ of order $p^{1+2n}$ and odd exponent $p$ are "good". To describe a "good" bijection I need some notation. Let $f_n\colon V\times V\to\mathbb{F}_p$ be a nondegenerate symplectic form on $V=\mathbb{F}_p^{2n}$. Multiplication in $G_n=V\times \mathbb{F}_p$ is given by $(v_1,\lambda_1)(v_2,\lambda_2)=(v_1+v_2,\lambda_1+\lambda_2+{\frac12}f_n(v_1,v_2))$, or by the matrices you indicate. The conjugacy classes are as follows: the $p$ one-element (central) classes $Z_\lambda:=\{(0,\lambda)\}$, $\lambda\in\mathbb{F}_p$, and the $p^{2n}-1$ classes $C_v:=\{(v,\lambda)\mid \lambda\in\mathbb{F}_p\}$ where $0\neq v\in V$. The irreducible representations are also easy. The trivial degree-1 representation corresponds to class $Z_0$ containing the identity element. The $p^{2n}-1$ nontrivial degree-1 representations correspond to the $p$-element classes $C_v$. The remaining $p-1$ irreducibles of degree $p^n$ correspond to the $p-1$ central conjugacy classes $Z_\lambda$, with $0\neq\lambda\in\mathbb{F}_p$. Fix a maximal totally isotropic subspace $W$ of $V$. By Witt's theorem $|W|=p^n$. Then $A:=W\times\mathbb{F}_p$ is a maximal abelian subgroup of $G_n$ of index $p^n$. Let $\sigma_\lambda$ be the 1-dimensional representation of $A$, with kernel $W$, mapping $(0,1)$ to $e^{2\pi i\lambda/p}$. The induced representations $\rho_\lambda={\rm Ind}_A^{G_n}(\sigma_\lambda)$ are irreducible of degree $p^n$. (A direct calculation shows $\langle\rho_\lambda,\rho_\lambda\rangle=1$. Choosing a different $f_n$, or a different maximal totally isotropic subspace $W'$, gives equivalent representations $\rho'_\lambda$. The $W$s are permuted by ${\rm Aut}(G_n)$.) This is a "good" bijection, as identifying $V$ with its dual seems allowed.

This is an interesting question, even though it is not well defined. Call a group "good" if it has a "good" bijection between its conjugacy classes and its irreducible complex representations. I agree with Alexander that the definition of a "good" bijection/group should be guided by classes of examples, but I prefer that a class of bijections/groups should be infinite.

There are families of good metacyclic groups. For example, if $n=2m+1$ is an odd integer, then the dihedral groups $D_{2n}=\langle a,b\mid a^2=b^n=1,\; b^{-1}ab=b^{-1}\rangle$ of order $2n$ are good. The conjugacy classes $\{b^j,b^{-j}\}$, $1\leq k\leq m$, $\{1\}$, and $\{a,ab,ab^2,\dots,ab^{n-1}\}$ correspond bijectively (I belive this is "good") to the irreducible representations $\rho_j$, $1\leq j\leq m$, $\sigma_0$, and $\sigma_1$, respectively where $\rho_j(a)=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\rho_j(b)=\begin{pmatrix}\zeta_n^j&0\\0&\zeta_n^{-j}\end{pmatrix}$, $\sigma_k(a)=\begin{pmatrix}(-1)^k\end{pmatrix}$, $\sigma_k(b)=\begin{pmatrix}1\end{pmatrix}$ and $\zeta_n=e^{2\pi i/n}$.

If an infinite family $G_1, G_2,\dots$ of groups is good, then you know a vast amount about each $G_n$ and can likely produce is a formula writing $|G_n|$ as a sum of the squares of the degrees of the irreducible representations. For $D_{2n}$ this is $2n=4m+2=m\times 2^2+2\times 1^2$. If $G_n$ is an extraspecial group of (odd) order $p^{1+2n}$ and exponent $p$, then the formula is $p^{1+2n}=(p-1)\times(p^n)^2+p^{2n}\times 1^2$. Perhaps the existence of such a formula should be part of the elusive definition of "good".

Addition: Yes Alexander, you are correct, the extraspecial group $G_n$ of order $p^{1+2n}$ and odd exponent $p$ is "good". To describe a "good" bijection I need some notation. Let $f_n\colon V\times V\to\mathbb{F}_p$ be a nondegenerate symplectic form on $V=\mathbb{F}_p^{2n}$. Multiplication in $G_n=V\times \mathbb{F}_p$ is given by $(v_1,\lambda_1)(v_2,\lambda_2)=(v_1+v_2,\lambda_1+\lambda_2+{\frac12}f_n(v_1,v_2))$, or by the matrices you indicate. The conjugacy classes are as follows: the $p$ one-element (central) classes $Z_\lambda:=\{(0,\lambda)\}$, $\lambda\in\mathbb{F}_p$, and the $p^{2n}-1$ classes $C_v:=\{(v,\lambda)\mid \lambda\in\mathbb{F}_p\}$ where $0\neq v\in V$. The irreducible representations are also easy. The trivial representation corresponds to the trivial character (or degree-1 representation). The $p^{2n}-1$ nontrivial characters correspond to the $p$-element classes $C_v$. The remaining $p-1$ irreducibles of degree $p^n$ correspond to the $p-1$ central conjugacy classes $Z_\lambda$, with $0\neq\lambda\in\mathbb{F}_p$. Fix a maximal totally isotropic subspace $W$ of $V$. By Witt's theorem $|W|=p^n$. Then $A:=W\times\mathbb{F}_p$ is a maximal abelian subgroup of $G_n$ of index $p^n$. Let $\sigma_\lambda$ be the 1-dimensional representation of $A$, with kernel $W$, mapping $(0,1)$ to $e^{2\pi i\lambda/p}$. The induced representations $\rho_\lambda={\rm Ind}_A^{G_n}(\sigma_\lambda)$ are irreducible of degree $p^n$. (A direct calculation shows $\langle\rho_\lambda,\rho_\lambda\rangle=1$. Choosing a different maximal totally isotropic subspace $W'$ gives equivalent representations $\rho'_\lambda$. The $W$s are permuted by ${\rm Aut}(G_n)$.) This is a "good" bijection, as identifying $V$ with its dual seems allowed.

This is an interesting question, even though it is not well defined. Call a group "good" if it has a "good" bijection between its conjugacy classes and its irreducible complex representations. I agree with Alexander that the definition of a "good" bijection/group should be guided by classes of examples, but I prefer that a class of bijections/groups should be infinite.

There are families of good metacyclic groups. For example, if $n=2m+1$ is an odd integer, then the dihedral groups $D_{2n}=\langle a,b\mid a^2=b^n=1,\; a^{-1}ba=b^{-1}\rangle$ of order $2n$ are good. The conjugacy classes $\{b^j,b^{-j}\}$, $1\leq j\leq m$, $\{1\}$, and $\{a,ab,ab^2,\dots,ab^{n-1}\}$ correspond bijectively (I believe this is "good") to the irreducible representations $\rho_j$, $1\leq j\leq m$, $\sigma_0$, and $\sigma_1$, respectively where $\rho_j(a)=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\rho_j(b)=\begin{pmatrix}\zeta_n^j&0\\0&\zeta_n^{-j}\end{pmatrix}$, $\sigma_k(a)=\begin{pmatrix}(-1)^k\end{pmatrix}$, $\sigma_k(b)=\begin{pmatrix}1\end{pmatrix}$ and $\zeta_n=e^{2\pi i/n}$.

If an infinite family $G_1, G_2,\dots$ of groups is good, then you know a vast amount about each $G_n$ and can likely produce is a formula writing $|G_n|$ as a sum of the squares of the degrees of the irreducible representations. For $D_{2n}$ this is $2n=4m+2=m\times 2^2+2\times 1^2$. If $G_n$ is an extraspecial group of (odd) order $p^{1+2n}$ and exponent $p$, then the formula is $p^{1+2n}=(p-1)\times(p^n)^2+p^{2n}\times 1^2$. Perhaps the existence of such a formula should be part of the elusive definition of "good".

Addition: Yes Alexander, you are correct, the extraspecial group $G_n$ of order $p^{1+2n}$ and odd exponent $p$ is "good". To describe a "good" bijection I need some notation. Let $f_n\colon V\times V\to\mathbb{F}_p$ be a nondegenerate symplectic form on $V=\mathbb{F}_p^{2n}$. Multiplication in $G_n=V\times \mathbb{F}_p$ is given by $(v_1,\lambda_1)(v_2,\lambda_2)=(v_1+v_2,\lambda_1+\lambda_2+{\frac12}f_n(v_1,v_2))$, or by the matrices you indicate. The conjugacy classes are as follows: the $p$ one-element (central) classes $Z_\lambda:=\{(0,\lambda)\}$, $\lambda\in\mathbb{F}_p$, and the $p^{2n}-1$ classes $C_v:=\{(v,\lambda)\mid \lambda\in\mathbb{F}_p\}$ where $0\neq v\in V$. The irreducible representations are also easy. The trivial representation corresponds to the trivial character (or degree-1 representation). The $p^{2n}-1$ nontrivial characters correspond to the $p$-element classes $C_v$. The remaining $p-1$ irreducibles of degree $p^n$ correspond to the $p-1$ central conjugacy classes $Z_\lambda$, with $0\neq\lambda\in\mathbb{F}_p$. Fix a maximal totally isotropic subspace $W$ of $V$. By Witt's theorem $|W|=p^n$. Then $A:=W\times\mathbb{F}_p$ is a maximal abelian subgroup of $G_n$ of index $p^n$. Let $\sigma_\lambda$ be the 1-dimensional representation of $A$, with kernel $W$, mapping $(0,1)$ to $e^{2\pi i\lambda/p}$. The induced representations $\rho_\lambda={\rm Ind}_A^{G_n}(\sigma_\lambda)$ are irreducible of degree $p^n$. (A direct calculation shows $\langle\rho_\lambda,\rho_\lambda\rangle=1$. Choosing a different maximal totally isotropic subspace $W'$ gives equivalent representations $\rho'_\lambda$. The $W$s are permuted by ${\rm Aut}(G_n)$.) This is a "good" bijection, as identifying $V$ with its dual seems allowed.

This is an interesting question, even though it is not well defined. Call a group "good" if it has a "good" bijection between its conjugacy classes and its irreducible complex representations. I agree with Alexander that the definition of a "good" bijection/group should be guided by classes of examples, but I prefer that a class of bijections/groups should be infinite.

There are families of good metacyclic groups. For example, if $n=2m+1$ is an odd integer, then the dihedral groups $D_{2n}=\langle a,b\mid a^2=b^n=1,\; b^{-1}ab=b^{-1}\rangle$ of order $2n$ are good. The conjugacy classes $\{b^j,b^{-j}\}$, $1\leq k\leq m$, $\{1\}$, and $\{a,ab,ab^2,\dots,ab^{n-1}\}$ correspond bijectively (I belive this is "good") to the irreducible representations $\rho_j$, $1\leq j\leq m$, $\sigma_0$, and $\sigma_1$, respectively where $\rho_j(a)=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\rho_j(b)=\begin{pmatrix}\zeta_n^j&0\\0&\zeta_n^{-j}\end{pmatrix}$, $\sigma_k(a)=\begin{pmatrix}(-1)^k\end{pmatrix}$, $\sigma_k(b)=\begin{pmatrix}1\end{pmatrix}$ and $\zeta_n=e^{2\pi i/n}$.

If an infinite family $G_1, G_2,\dots$ of groups is good, then you know a vast amount about each $G_n$ and can likely produce is a formula writing $|G_n|$ as a sum of the squares of the degrees of the irreducible representations. For $D_{2n}$ this is $2n=4m+2=m\times 2^2+2\times 1^2$. If $G_n$ is an extraspecial group of (odd) order $p^{1+2n}$ and exponent $p$, then the formula is $p^{1+2n}=(p-1)\times(p^n)^2+(p^{2n}-1)\times 1^2$. Perhaps the existence of such a formula should be part of the elusive definition of "good".

This is an interesting question, even though it is not well defined. Call a group "good" if it has a "good" bijection between its conjugacy classes and its irreducible complex representations. I agree with Alexander that the definition of a "good" bijection/group should be guided by classes of examples, but I prefer that a class of bijections/groups should be infinite.

There are families of good metacyclic groups. For example, if $n=2m+1$ is an odd integer, then the dihedral groups $D_{2n}=\langle a,b\mid a^2=b^n=1,\; b^{-1}ab=b^{-1}\rangle$ of order $2n$ are good. The conjugacy classes $\{b^j,b^{-j}\}$, $1\leq k\leq m$, $\{1\}$, and $\{a,ab,ab^2,\dots,ab^{n-1}\}$ correspond bijectively (I belive this is "good") to the irreducible representations $\rho_j$, $1\leq j\leq m$, $\sigma_0$, and $\sigma_1$, respectively where $\rho_j(a)=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\rho_j(b)=\begin{pmatrix}\zeta_n^j&0\\0&\zeta_n^{-j}\end{pmatrix}$, $\sigma_k(a)=\begin{pmatrix}(-1)^k\end{pmatrix}$, $\sigma_k(b)=\begin{pmatrix}1\end{pmatrix}$ and $\zeta_n=e^{2\pi i/n}$.

If an infinite family $G_1, G_2,\dots$ of groups is good, then you know a vast amount about each $G_n$ and can likely produce is a formula writing $|G_n|$ as a sum of the squares of the degrees of the irreducible representations. For $D_{2n}$ this is $2n=4m+2=m\times 2^2+2\times 1^2$. If $G_n$ is an extraspecial group of (odd) order $p^{1+2n}$ and exponent $p$, then the formula is $p^{1+2n}=(p-1)\times(p^n)^2+p^{2n}\times 1^2$. Perhaps the existence of such a formula should be part of the elusive definition of "good".

Addition: Yes Alexander, you are correct, the extraspecial group $G_n$ of order $p^{1+2n}$ and odd exponent $p$ is "good". To describe a "good" bijection I need some notation. Let $f_n\colon V\times V\to\mathbb{F}_p$ be a nondegenerate symplectic form on $V=\mathbb{F}_p^{2n}$. Multiplication in $G_n=V\times \mathbb{F}_p$ is given by $(v_1,\lambda_1)(v_2,\lambda_2)=(v_1+v_2,\lambda_1+\lambda_2+{\frac12}f_n(v_1,v_2))$, or by the matrices you indicate. The conjugacy classes are as follows: the $p$ one-element (central) classes $Z_\lambda:=\{(0,\lambda)\}$, $\lambda\in\mathbb{F}_p$, and the $p^{2n}-1$ classes $C_v:=\{(v,\lambda)\mid \lambda\in\mathbb{F}_p\}$ where $0\neq v\in V$. The irreducible representations are also easy. The trivial representation corresponds to the trivial character (or degree-1 representation). The $p^{2n}-1$ nontrivial characters correspond to the $p$-element classes $C_v$. The remaining $p-1$ irreducibles of degree $p^n$ correspond to the $p-1$ central conjugacy classes $Z_\lambda$, with $0\neq\lambda\in\mathbb{F}_p$. Fix a maximal totally isotropic subspace $W$ of $V$. By Witt's theorem $|W|=p^n$. Then $A:=W\times\mathbb{F}_p$ is a maximal abelian subgroup of $G_n$ of index $p^n$. Let $\sigma_\lambda$ be the 1-dimensional representation of $A$, with kernel $W$, mapping $(0,1)$ to $e^{2\pi i\lambda/p}$. The induced representations $\rho_\lambda={\rm Ind}_A^{G_n}(\sigma_\lambda)$ are irreducible of degree $p^n$. (A direct calculation shows $\langle\rho_\lambda,\rho_\lambda\rangle=1$. Choosing a different maximal totally isotropic subspace $W'$ gives equivalent representations $\rho'_\lambda$. The $W$s are permuted by ${\rm Aut}(G_n)$.) This is a "good" bijection, as identifying $V$ with its dual seems allowed.