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This is, of course, a well-known fact (though the inequality need not be strict), but here's one proof. Let $H$ be a finite group, $Z$ be a central subgroup, and $\lambda$ be a linear character of $Z.$ Then the number of irreducible characters of $H$ which lie over $\lambda$ is $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)}.$ Now for each $z \in Z$, the quantity $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi (z)}{\chi(1)}$ is non-negative,as was probably known to Burnside, and is non-zero if and only if $z$ is a commutator in $H$. It follows easily that for any irreducible character $\lambda$ of $Z,$ we have $ \sum_{z \in Z} \left( \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \right) \lambda(z^{-1}) \leq \sum_{z \in Z}\sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} = |Z| k(H/Z),$ where $k(X)$ denotes the number of conjugacy classes of $X.$ The last equality follows because the irreducible characters of $H$ with $Z$ in their kernels are precisely those which contain the trivial character on restriction to $Z.$ Hence $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)} \leq k(H/Z),$ as claimed. Furthermore, if there is a non-identity element $z \in Z \backslash {\rm ker}(\lambda)$ which is a commutator in $H$, then the inequality becomes strict because $\lambda(z^{-1})$ has real part less than $1$. If there is no such commutator $z$, then we have $ \left( \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \right) \lambda(z^{-1}) = \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} $ for each $z \in Z$ and then $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)} = k(H/Z).$

Note that given a factor set $\alpha$ taking values in roots of unity and a finite group ${\tilde H},$ we obtain a finite group $H$ with $H/Z \cong {\tilde H},$ where $Z$ is a central subgroup.

Since the OP could not find a reference, here is a proof of the inequality of Burnside. Note that an element $x \in G$ is a (simple) commutator if and only if $x = y^{-1}z$ for conjugate elements $y,z \in G.$ Now for $x \in G,$ the number of times that $x$ may be expressed in the form $a^{-1}b$ with $a,b$ both conjugate to $y$ is given by the class algebra constant formula $\frac{|G|}{|C_{G}(y)|^{2}} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x^{-1})}{\chi(1)}$. This is clearly a non-negative integer, and is certainly real. We may take the complex conjugate, and multiply through by $|C_{G}(y)|$ to deduce that $\frac{|G|}{|C_{G}(y)|} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x)}{\chi(1)}$ is real and non-negative. Summing over a set of representatives $y$ for the conjugacy classes of $G,$ we may conclude that $|G| \sum_{\chi \in {\rm Irr}(G)}\frac{\chi(x)}{\chi(1)}$ is real and non-negative, and is strictly positive if and only if $x$ is a commutator.

This is, of course, a well-known fact (though the inequality need not be strict), but here's one proof. Let $H$ be a finite group, $Z$ be a central subgroup, and $\lambda$ be a linear character of $Z.$ Then the number of irreducible characters of $H$ which lie over $\lambda$ is $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)}.$ Now for each $z \in Z$, the quantity $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi (z)}{\chi(1)}$ is non-negative,as was probably known to Burnside, and is non-zero if and only if $z$ is a commutator in $H$ (which certainly was known to Burnside, though set as an exercise in his book). It follows easily that for any irreducible character $\lambda$ of $Z,$ we have $ \sum_{z \in Z} \left( \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \right) \lambda(z^{-1}) \leq \sum_{z \in Z}\sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} = |Z| k(H/Z),$ where $k(X)$ denotes the number of conjugacy classes of $X.$ The last equality follows because the irreducible characters of $H$ with $Z$ in their kernels are precisely those which contain the trivial character on restriction to $Z.$ Hence $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)} \leq k(H/Z),$ as claimed. Furthermore, if there is a non-identity element $z \in Z \backslash {\rm ker}(\lambda)$ which is a commutator in $H$, then the inequality becomes strict because $\lambda(z^{-1})$ has real part less than $1$. If there is no such commutator $z$, then we have $ \left( \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \right) \lambda(z^{-1}) = \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} $ for each $z \in Z$ and then $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)} = k(H/Z).$

Note that given a factor set $\alpha$ taking values in roots of unity and associated to a finite group ${\tilde H},$ we obtain a finite group $H$ with $H/Z \cong {\tilde H},$ where $Z$ is a central subgroup.

Since the OP could not find a reference, here is a proof of the inequality of Burnside. Note that an element $x \in G$ is a (simple) commutator if and only if $x = y^{-1}z$ for conjugate elements $y,z \in G.$ Now for $x \in G,$ the number of times that $x$ may be expressed in the form $a^{-1}b$ with $a,b$ both conjugate to $y$ is given by the class algebra constant formula $\frac{|G|}{|C_{G}(y)|^{2}} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x^{-1})}{\chi(1)}$. This is clearly a non-negative integer, and is certainly real. We may take the complex conjugate, and multiply through by $|C_{G}(y)|$ to deduce that $\frac{|G|}{|C_{G}(y)|} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x)}{\chi(1)}$ is real and non-negative. Summing over a set of representatives $y$ for the conjugacy classes of $G,$ we may conclude that $|G| \sum_{\chi \in {\rm Irr}(G)}\frac{\chi(x)}{\chi(1)}$ is real and non-negative, and is strictly positive if and only if $x$ is a commutator.

This is, of course, a well-known fact (though the inequality need not be strict), but here's one proof. Let $H$ be a finite group, $Z$ be s central subgroup, and $\lambda$ be a linear character of $Z.$ Then the number of irreducible characters of $H$ which lie over $\lambda$ is $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)}.$ Now for each $z \in Z(,$ the quantity $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi (z)}{\chi(1)}$ was shown by Burnside to be non-negative, and non-zero if and only if $z$ is a commutator in $H$. It follows easily that for any irreducible character $\lambda$ of $Z,$ we have$ \sum_{z \in Z} \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \lambda(z^{-1}) \leq \sum_{z \in Z}\sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} = |Z| k(H/Z),$ where $k(X)$ denotes the number of conjugacy classes of $X.$ The last equality follows because the irreducible characters of $H$ with $Z$ in their kernels are precisely those which contain the trivial character on restriction to $Z.$ Hence $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)} \leq k(H/Z),$ as claimed. Furthermore, if (and only if) there is a non-identity element $z \in Z \backslash {\rm ker}(\lambda)$ which is a commutator in $H$, then the inequality becomes strict.

Note that given a factor set $\alpha$ taking values in roots of unity and a finite group ${\tilde H},$ we obtain a finite group $H$ with $H/Z \cong {\tilde H},$ where $Z$ is a central subgroup.

Since the OP could not find a reference, here is a proof of the inequality of Burnside. Note that an element $x \in G$ is a (simple) commutator if and only if $x = y^{-1}z$ for conjugate elements $y,z \in G.$ Now for $x \in G,$ the number of times that $x$ may be expressed in the form $a^{-1}b$ with $a,b$ both conjugate to $y$ is given by the class algebra constant formula $\frac{|G|}{|C_{G}(y)|^{2}} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x^{-1})}{\chi(1)}$. This is clearly a non-negative integer, and is certainly real. We may take the complex conjugate, and multiply through by $|C_{G}(y)|$ to deduce that $\frac{|G|}{|C_{G}(y)|} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x)}{\chi(1)}$ is real and non-negative. Summing over a set of representatives $y$ for the conjugacy classes of $G,$ we may conclude that $|G| \sum_{\chi \in {\rm Irr}(G)}\frac{\chi(x)}{\chi(1)}$ is real and non-negative, and is strictly positive if and only if $x$ is a commutator.

This is, of course, a well-known fact (though the inequality need not be strict), but here's one proof. Let $H$ be a finite group, $Z$ be a central subgroup, and $\lambda$ be a linear character of $Z.$ Then the number of irreducible characters of $H$ which lie over $\lambda$ is $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)}.$ Now for each $z \in Z$, the quantity $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi (z)}{\chi(1)}$ is non-negative,as was probably known to Burnside, and is non-zero if and only if $z$ is a commutator in $H$. It follows easily that for any irreducible character $\lambda$ of $Z,$ we have $ \sum_{z \in Z} \left( \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \right) \lambda(z^{-1}) \leq \sum_{z \in Z}\sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} = |Z| k(H/Z),$ where $k(X)$ denotes the number of conjugacy classes of $X.$ The last equality follows because the irreducible characters of $H$ with $Z$ in their kernels are precisely those which contain the trivial character on restriction to $Z.$ Hence $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)} \leq k(H/Z),$ as claimed. Furthermore, if there is a non-identity element $z \in Z \backslash {\rm ker}(\lambda)$ which is a commutator in $H$, then the inequality becomes strict because $\lambda(z^{-1})$ has real part less than $1$. If there is no such commutator $z$, then we have $ \left( \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \right) \lambda(z^{-1}) = \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} $ for each $z \in Z$ and then $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)} = k(H/Z).$

Note that given a factor set $\alpha$ taking values in roots of unity and a finite group ${\tilde H},$ we obtain a finite group $H$ with $H/Z \cong {\tilde H},$ where $Z$ is a central subgroup.

Since the OP could not find a reference, here is a proof of the inequality of Burnside. Note that an element $x \in G$ is a (simple) commutator if and only if $x = y^{-1}z$ for conjugate elements $y,z \in G.$ Now for $x \in G,$ the number of times that $x$ may be expressed in the form $a^{-1}b$ with $a,b$ both conjugate to $y$ is given by the class algebra constant formula $\frac{|G|}{|C_{G}(y)|^{2}} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x^{-1})}{\chi(1)}$. This is clearly a non-negative integer, and is certainly real. We may take the complex conjugate, and multiply through by $|C_{G}(y)|$ to deduce that $\frac{|G|}{|C_{G}(y)|} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x)}{\chi(1)}$ is real and non-negative. Summing over a set of representatives $y$ for the conjugacy classes of $G,$ we may conclude that $|G| \sum_{\chi \in {\rm Irr}(G)}\frac{\chi(x)}{\chi(1)}$ is real and non-negative, and is strictly positive if and only if $x$ is a commutator.

This is, of course, a well-known fact, but here's one proof. Let $H$ be a finite group, $Z$ be s central subgroup, and $\lambda$ be a linear character of $Z.$ Then the number of irreducible characters of $H$ which lie over $\lambda$ is $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)}.$ Now for each $z \in Z(,$ the quantity $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi (z)}{\chi(1)}$ was shown by Burnside to be non-negative, and non-zero if and only if $z$ is a commutator in $H$. It follows easily that for any irreducible character $\lambda$ of $Z,$ we have$ \sum_{z \in Z} \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \lambda(z^{-1}) \leq \sum_{z \in Z}\sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} = |Z| k(H/Z),$ where $k(X)$ denotes the number of conjugacy classes of $X.$ The last equality follows because the irreducible characters of $H$ with $Z$ in their kernels are precisely those which contain the trivial character on restriction to $Z.$ Hence $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)} \leq k(H/Z),$ as claimed. Furthermore, if (and only if) there is a non-identity element $z \in Z \backslash {\rm ker}(\lambda)$ which is a commutator in $H$, then the inequality becomes strict.

Note that given a factor set $\alpha$ taking values in roots of unity and a finite group ${\tilde H},$ we obtain a finite group $H$ with $H/Z \cong {\tilde H},$ where $Z$ is a central subgroup.

Since the OP could not find a reference, here is a proof of the inequality of Burnside. Note that an element $x \in G$ is a (simple) commutator if and only if $x = y^{-1}z$ for conjugate elements $y,z \in G.$ Now for $x \in G,$ the number of times that $x$ may be expressed in the form $a^{-1}b$ with $a,b$ both conjugate to $y$ is given by the class algebra constant formula $\frac{|G|}{|C_{G}(y)|^{2}} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x^{-1})}{\chi(1)}$. This is clearly a non-negative integer, and is certainly real. We may take the complex conjugate, and multiply through by $|C_{G}(y)|$ to deduce that $\frac{|G|}{|C_{G}(y)|} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x)}{\chi(1)}$ is real and non-negative. Summing over a set of representatives $y$ for the conjugacy classes of $G,$ we may conclude that $|G| \sum_{\chi \in {\rm Irr}(G)}\frac{\chi(x)}{\chi(1)}$ is real and non-negative, and is strictly positive if and only if $x$ is a commutator.

This is, of course, a well-known fact (though the inequality need not be strict), but here's one proof. Let $H$ be a finite group, $Z$ be s central subgroup, and $\lambda$ be a linear character of $Z.$ Then the number of irreducible characters of $H$ which lie over $\lambda$ is $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)}.$ Now for each $z \in Z(,$ the quantity $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi (z)}{\chi(1)}$ was shown by Burnside to be non-negative, and non-zero if and only if $z$ is a commutator in $H$. It follows easily that for any irreducible character $\lambda$ of $Z,$ we have$ \sum_{z \in Z} \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \lambda(z^{-1}) \leq \sum_{z \in Z}\sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} = |Z| k(H/Z),$ where $k(X)$ denotes the number of conjugacy classes of $X.$ The last equality follows because the irreducible characters of $H$ with $Z$ in their kernels are precisely those which contain the trivial character on restriction to $Z.$ Hence $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z}(\chi), \lambda \rangle}{\chi(1)} \leq k(H/Z),$ as claimed. Furthermore, if (and only if) there is a non-identity element $z \in Z \backslash {\rm ker}(\lambda)$ which is a commutator in $H$, then the inequality becomes strict.

Note that given a factor set $\alpha$ taking values in roots of unity and a finite group ${\tilde H},$ we obtain a finite group $H$ with $H/Z \cong {\tilde H},$ where $Z$ is a central subgroup.

Since the OP could not find a reference, here is a proof of the inequality of Burnside. Note that an element $x \in G$ is a (simple) commutator if and only if $x = y^{-1}z$ for conjugate elements $y,z \in G.$ Now for $x \in G,$ the number of times that $x$ may be expressed in the form $a^{-1}b$ with $a,b$ both conjugate to $y$ is given by the class algebra constant formula $\frac{|G|}{|C_{G}(y)|^{2}} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x^{-1})}{\chi(1)}$. This is clearly a non-negative integer, and is certainly real. We may take the complex conjugate, and multiply through by $|C_{G}(y)|$ to deduce that $\frac{|G|}{|C_{G}(y)|} \sum_{\chi \in {\rm Irr}(G)}\frac{|\chi(y)|^{2} \chi(x)}{\chi(1)}$ is real and non-negative. Summing over a set of representatives $y$ for the conjugacy classes of $G,$ we may conclude that $|G| \sum_{\chi \in {\rm Irr}(G)}\frac{\chi(x)}{\chi(1)}$ is real and non-negative, and is strictly positive if and only if $x$ is a commutator.