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Revised: The answer to 1 is a qualified "yes", though as Frieder Ladisch points out, it might be more accurate to say that the number of squares can be determined by (irreducible) character-theoretic information. The number of square roots of $g \in G$ ( $G$ a finite group) is given by $\sum_{\chi \in {\rm Irr}(G)} \nu(\chi) \chi(g)$, where $\nu(\chi)$ denotes the Frobenius-Schur indicator (and $\chi$ runs over all complex irreducible characters of $G$). This formula is well-known (though not, perhaps, as well-known as the special case $g = 1_{G}$). It is a direct consequence of the orthogonality relations and the fact that $\nu(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^{2})$. Hence those $g \in G$ which have a square root are known from the character table, and the number of such is then known by the orthogonality relations.

Revised again: Returning to Frieder's point and question, it is possible that, although the F-S indicator is not in general determinable from the character table alone, the number of squares in the group still might be ( I do not assert that that IS the case- I just do not know at present- NOW SETTLED (in the negative) in view of Frieder's answer). It might be of interest to note that if we label so that $1 = \chi_{1}, \chi_{2},\ldots \chi_{r}$ are all the real-valued irreducible characters of $G$ ( these can obviously be identified from the character table), then we know because of the F-S indicator that there is a choice of signs $\varepsilon_{1},\ldots,\varepsilon_{r}$ such that the class function $\sum_{i=1}^{r} \varepsilon_{i}\chi_{i}$ assumes non-negative integer values everywhere - we could ask whether there is more than one choice of signs with this non-negativity property: I also find it interesting (though elementary) to note that the "expected number" of square roots of an element of any finite group $G$ is $1$. The last statement is just saying that if we sum the number of square roots of $g$ over all $g \in G$, we obtain $|G|$, because the "square-root" counting function $\sum_{\chi} \nu(\chi)\chi$ contains the trivial character with multiplicity one. LATER NOTE: While the expected number of square roots of $g \in G$ is $1$, the variance of the distribution is easily checked to be $\sqrt{k(G)-1}$, where $k(G)$ is the number of conjugacy classes of $G$.

I do not know the answer to 2, though others might.

Revised: The answer to 1 is a qualified "yes", though as Frieder Ladisch points out, it might be more accurate to say that the number of squares can be determined by (irreducible) character-theoretic information. The number of square roots of $g \in G$ ( $G$ a finite group) is given by $\sum_{\chi \in {\rm Irr}(G)} \nu(\chi) \chi(g)$, where $\nu(\chi)$ denotes the Frobenius-Schur indicator (and $\chi$ runs over all complex irreducible characters of $G$). This formula is well-known (though not, perhaps, as well-known as the special case $g = 1_{G}$). It is a direct consequence of the orthogonality relations and the fact that $\nu(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^{2})$. Hence those $g \in G$ which have a square root are known from the character table, and the number of such is then known by the orthogonality relations.

Revised again: Returning to Frieder's point and question, it is possible that, although the F-S indicator is not in general determinable from the character table alone, the number of squares in the group still might be ( I do not assert that that IS the case- I just do not know at present- NOW SETTLED (in the negative) in view of Frieder's answer). It might be of interest to note that if we label so that $1 = \chi_{1}, \chi_{2},\ldots \chi_{r}$ are all the real-valued irreducible characters of $G$ ( these can obviously be identified from the character table), then we know because of the F-S indicator that there is a choice of signs $\varepsilon_{1},\ldots,\varepsilon_{r}$ such that the class function $\sum_{i=1}^{r} \varepsilon_{i}\chi_{i}$ assumes non-negative integer values everywhere - we could ask whether there is more than one choice of signs with this non-negativity property: I also find it interesting (though elementary) to note that the "expected number" of square roots of an element of any finite group $G$ is $1$. The last statement is just saying that if we sum the number of square roots of $g$ over all $g \in G$, we obtain $|G|$, because the "square-root" counting function $\sum_{\chi} \nu(\chi)\chi$ contains the trivial character with multiplicity one. LATER NOTE: While the expected number of square roots of $g \in G$ is $1$, the variance of the distribution is easily checked to be $k(G)-1$, where $k(G)$ is the number of conjugacy classes of $G$.

I do not know the answer to 2, though others might.

Revised: The answer to 1 is a qualified "yes", though as Frieder Ladisch points out, it might be more accurate to say that the number of squares can be determined by (irreducible) character-theoretic information. The number of square roots of $g \in G$ ( $G$ a finite group) is given by $\sum_{\chi \in {\rm Irr}(G)} \nu(\chi) \chi(g)$, where $\nu(\chi)$ denotes the Frobenius-Schur indicator (and $\chi$ runs over all complex irreducible characters of $G$). This formula is well-known (though not, perhaps, as well-known as the special case $g = 1_{G}$). It is a direct consequence of the orthogonality relations and the fact that $\nu(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^{2})$. Hence those $g \in G$ which have a square root are known from the character table, and the number of such is then known by the orthogonality relations.

Revised again: Returning to Frieder's point and question, it is possible that, although the F-S indicator is not in general determinable from the character table alone, the number of squares in the group still might be ( I do not assert that that IS the case- I just do not know at present- NOW SETTLED (in the negative) in view of Frieder's answer). It might be of interest to note that if we label so that $1 = \chi_{1}, \chi_{2},\ldots \chi_{r}$ are all the real-valued irreducible characters of $G$ ( these can obviously be identified from the character table), then we know because of the F-S indicator that there is a choice of signs $\varepsilon_{1},\ldots,\varepsilon_{r}$ such that the class function $\sum_{i=1}^{r} \varepsilon_{i}\chi_{i}$ assumes non-negative integer values everywhere - we could ask whether there is more than one choice of signs with this non-negativity property: I also find it interesting (though elementary) to note that the "expected number" of square roots of an element of any finite group $G$ is $1$. The last statement is just saying that if we sum the number of square roots of $g$ over all $g \in G$, we obtain $|G|$, because the "square-root" counting function $\sum_{\chi} \nu(\chi)\chi$ contains the trivial character with multiplicity one.

I do not know the answer to 2, though others might.

Revised: The answer to 1 is a qualified "yes", though as Frieder Ladisch points out, it might be more accurate to say that the number of squares can be determined by (irreducible) character-theoretic information. The number of square roots of $g \in G$ ( $G$ a finite group) is given by $\sum_{\chi \in {\rm Irr}(G)} \nu(\chi) \chi(g)$, where $\nu(\chi)$ denotes the Frobenius-Schur indicator (and $\chi$ runs over all complex irreducible characters of $G$). This formula is well-known (though not, perhaps, as well-known as the special case $g = 1_{G}$). It is a direct consequence of the orthogonality relations and the fact that $\nu(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^{2})$. Hence those $g \in G$ which have a square root are known from the character table, and the number of such is then known by the orthogonality relations.

Revised again: Returning to Frieder's point and question, it is possible that, although the F-S indicator is not in general determinable from the character table alone, the number of squares in the group still might be ( I do not assert that that IS the case- I just do not know at present- NOW SETTLED (in the negative) in view of Frieder's answer). It might be of interest to note that if we label so that $1 = \chi_{1}, \chi_{2},\ldots \chi_{r}$ are all the real-valued irreducible characters of $G$ ( these can obviously be identified from the character table), then we know because of the F-S indicator that there is a choice of signs $\varepsilon_{1},\ldots,\varepsilon_{r}$ such that the class function $\sum_{i=1}^{r} \varepsilon_{i}\chi_{i}$ assumes non-negative integer values everywhere - we could ask whether there is more than one choice of signs with this non-negativity property: I also find it interesting (though elementary) to note that the "expected number" of square roots of an element of any finite group $G$ is $1$. The last statement is just saying that if we sum the number of square roots of $g$ over all $g \in G$, we obtain $|G|$, because the "square-root" counting function $\sum_{\chi} \nu(\chi)\chi$ contains the trivial character with multiplicity one. LATER NOTE: While the expected number of square roots of $g \in G$ is $1$, the variance of the distribution is easily checked to be $\sqrt{k(G)-1}$, where $k(G)$ is the number of conjugacy classes of $G$.

I do not know the answer to 2, though others might.

Revised: The answer to 1 is a qualified "yes", though as Frieder Ladisch points out, it might be more accurate too say that the number of squares can be determined by (irreducible) character-theoretic information. The number of square roots of $g \in G$ ( $G$ a finite group) is given by $\sum_{\chi \in {\rm Irr}(G)} \nu(\chi) \chi(g)$, where $\nu(\chi)$ denotes the Frobenius-Schur indicator (and $\chi$ runs over all complex irreducible characters of $G$). This formula is well-known (though not, perhaps, as well-known as the special case $g = 1_{G}$). It is a direct consequence of the orthogonality relations and the fact that $\nu(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^{2})$. Hence those $g \in G$ which have a square root are known from the character table, and the number of such is then known by the orthogonality relations.

Revised again: Returning to Frieder's point and question, it is possible that, although the F-S indicator is not in general determinable from the character table alone, the number of squares in the group still might be ( I do not assert that that IS the case- I just do not know at present- NOW SETTLED (in the negative) in view of Frieder's answer). It might be of interest to note that if we label so that $1 = \chi_{1}, \chi_{2},\ldots \chi_{r}$ are all the real-valued irreducible characters of $G$ ( these can obviously be identified from the character table), then we know because of the F-S indicator that there is a choice of signs $\varepsilon_{1},\ldots,\varepsilon_{r}$ such that the class function $\sum_{i=1}^{r} \varepsilon_{i}\chi_{i}$ assumes non-negative integer values everywhere - we could ask whether there is more than one choice of signs with this non-negativity property: I also find it interesting (though elementary) to note that the "expected number" of square roots of an element of any finite group $G$ is $1$. The last statement is just saying that if we sum the number of square roots of $g$ over all $g \in G$, we obtain $|G|$, because the "square-root" counting function $\sum_{\chi} \nu(\chi)\chi$ contains the trivial character with multiplicity one.

I do not know the answer to 2, though others might.

Revised: The answer to 1 is a qualified "yes", though as Frieder Ladisch points out, it might be more accurate to say that the number of squares can be determined by (irreducible) character-theoretic information. The number of square roots of $g \in G$ ( $G$ a finite group) is given by $\sum_{\chi \in {\rm Irr}(G)} \nu(\chi) \chi(g)$, where $\nu(\chi)$ denotes the Frobenius-Schur indicator (and $\chi$ runs over all complex irreducible characters of $G$). This formula is well-known (though not, perhaps, as well-known as the special case $g = 1_{G}$). It is a direct consequence of the orthogonality relations and the fact that $\nu(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^{2})$. Hence those $g \in G$ which have a square root are known from the character table, and the number of such is then known by the orthogonality relations.

Revised again: Returning to Frieder's point and question, it is possible that, although the F-S indicator is not in general determinable from the character table alone, the number of squares in the group still might be ( I do not assert that that IS the case- I just do not know at present- NOW SETTLED (in the negative) in view of Frieder's answer). It might be of interest to note that if we label so that $1 = \chi_{1}, \chi_{2},\ldots \chi_{r}$ are all the real-valued irreducible characters of $G$ ( these can obviously be identified from the character table), then we know because of the F-S indicator that there is a choice of signs $\varepsilon_{1},\ldots,\varepsilon_{r}$ such that the class function $\sum_{i=1}^{r} \varepsilon_{i}\chi_{i}$ assumes non-negative integer values everywhere - we could ask whether there is more than one choice of signs with this non-negativity property: I also find it interesting (though elementary) to note that the "expected number" of square roots of an element of any finite group $G$ is $1$. The last statement is just saying that if we sum the number of square roots of $g$ over all $g \in G$, we obtain $|G|$, because the "square-root" counting function $\sum_{\chi} \nu(\chi)\chi$ contains the trivial character with multiplicity one.

I do not know the answer to 2, though others might.