This was asked at MSE but never answered.

Let $G$ be a finite group and denote by $sq(G)$ the number of squares in $G$ i.e. the number of elements in $G$ which possess a square root. For example, if $G$ is a group of odd order, then $sq(G) =|G|$, since each element has a square root, while at the opposite extreme, $sq(G)= 1$ when $G$ is an elementary abelian 2-group. For the symmetric groups, the values of $sq(\mathrm{Sym}(n))$ are listed at $\,$https://oeis.org/A003483 $\,$. Finally, we note that both the dihedral and quaternion groups of order 8 share the value $sq(G)=2$.


1) Can $sq(G)$ be determined from information in the character table of $G$?

2) Is it true that $sq(\mathrm{Sym}(n))$ is divisible by every prime which is less than or equal to $n$ for $n>3$?


4 Answers 4


I answer here positively the second question (it's completely independent of the first one so it could have been 2 distinct posts).

Let $p\le n$ be prime. Let $C\subset S_n$ be the group generated by the cycle $(1\dots p)$. Let $C$ act on $S_n$ by conjugation. It preserves the set $Q$ of squares. Let $Q^C$ be the $C$-fixed points (that is, the set of squares centralizing $C$). The $C$-action on $Q\smallsetminus Q^C$ is free, hence $Q\smallsetminus Q^C$ has cardinality divisible by $p$. The centralizer $Z_C$ of $C$ is $C\times S_{n-p}$ (where $S_{n-p}$ is the pointwise stabilizer of $\{1,\dots,p\}$). Thus $Q^C=Z_C\cap Q$ by definition. If $Q(Z_C)$ is the set of squares of $Z_C$, then clearly $Q(Z_C)\subset Q^C$.

Actually if $p$ is odd, this is an equality (as noticed by "GH from MO", this equality is not a tautology and has to be checked). Indeed, if we have an element in $Q^C=Z_C\cap Q$, its number of $k$-cycles when $k$ is even is the same as its restriction to $\{p+1,\dots,n\}$ (since the possible $p$-cycle does not matter). So $Q^C=C\times\mathrm{Sq}(S_{n-p})$. Hence $\#(Q^C)=p\mathrm{sq}(n-p)$. So $Q$ has cardinality divisible by $p$.

If $p=2$ the previous argument fails at two points, since then $\mathrm{Sq}(Z_C)=\{1\}\times\mathrm{Sq}(S_{n-2})$ (instead of $C\times\mathrm{Sq}(S_{n-2})$), and since the cycle counting argument falls apart.

So we assume $n\ge 4$ and now redefine $C$ as being generated by the 4-cycle $(1\dots 2^k)$ with $k$ maximal ($k=\lfloor\log_2 n\rfloor$). Then $Q\smallsetminus Q^C$ has even cardinality, and the centralizer of $C$ is $C^2\times S_{n-4}$. We have $Q(Z_C)=C^2\times\mathrm{Sq}(S_{n-4})$ (here $C^2$ means the set of squares in $C$, which is a subgroup of order 2). Also $Q(Z_C)\subset Q^C$ and we need to check that it is an equality.

If some element $w$ belongs to $Q^C=Q\cap Z_C$ and $m$ is even, then its number $n_m=n_m(w)$ of $m$-cycles is even by assumption. Write $w=uv$ according to the decomposition $Z_C=C\times S_{n-2^k}$. If $m$ is not a power of $2$, then $u$ has no $m$-cycle, so $n_m(v)=n_m$. If $m$ is a power of 2 and $m<2^k$, then the number of $m$-cycles in $u$ is even (equal to either $2^k/m$ or 0), so $n_m(v)$ is even as well. If $m\ge 2^k$ we have a contradiction because $n-2^k<2^k$, so the only possibility is that $m=2^k$ and the $m$-cycle is supported by $\{1,\dots,2^k\}$, but then $n_m(w)=1$ and $w$ is not a square. Thus $u$ and $v$ are both squares in $C$ and $S_{n-2^k}$ respectively. So $Q^C=C^2\times\mathrm{Sq}(S_{n-4})$, which has even cardinality and thus $Q$ also has even cardinality. [Edit: I modified by initial argument for $p=2$ in which I supposed $k=2$ instead of $k$ maximal, but then $Q(Z_C)\subset Q^C$ can be strict as we see if $n\ge 8$ by taking the element $(1234)(5678)$. So the point noticed by "GH from MO" is not insignificant.]

  • 2
    $\begingroup$ I think it should be remarked (in the case $p>2$) that an element of $Q^C$ is automatically a square within the centralizer of $C$. This is not obvious, because a square-root of an element from $Q^C$ can be outside the centralizer of $C$, e.g. a $2p$-cycle of the form $(1\tau_1\dots p\tau_p)$ is a square-root of $(1\dots p)(\tau_1\dots\tau_p)$. $\endgroup$
    – GH from MO
    Oct 17, 2015 at 21:21
  • 2
    $\begingroup$ @GHfromMO Thanks very much. For $p$ odd this is indeed a simple verification, but for $p=2$ I had to modify the argument in a quite unexpected way. $\endgroup$
    – YCor
    Oct 17, 2015 at 22:56
  • 2
    $\begingroup$ It might be worth remarking that it is not generally true that in a finite group $G$ of even order, the number of squares is even (so there is indeed something to do for $S_{n}$): for example, in a group $G$ which has Sylow $2$-subgroups of exponent $2$, only elements of odd order are squares, and the number of elements of odd order in any finite group $G$ is odd ( as all elements of odd order except the identity occur in mutually inverse pairs). $\endgroup$ Oct 18, 2015 at 11:27
  • 2
    $\begingroup$ @GeoffRobinson Well, $S_3$ or even the cyclic group of order 6 being counterexamples, I think nobody conjectured such a fact :) $\endgroup$
    – YCor
    Oct 18, 2015 at 11:42

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.

  • 5
    $\begingroup$ In a very strict sense, we cannot compute the Frobenius-Schur indicator from the character table alone, since we can not determine the conjugacy class containing the squares of the elements in a given conjugacy class, if we have only the character table. An example would be the dihedral group and the quaternion group of order $8$. I wonder if the answer to 1) is "Yes" in this more strict sense? $\endgroup$ Oct 17, 2015 at 18:17
  • $\begingroup$ I have slightly revised my answer to take account of your useful comment. $\endgroup$ Oct 18, 2015 at 11:19
  • $\begingroup$ I deleted the other comment, too. By the way, isn't the last statement on the "expected number of squares" essentially clear, because every $g\in G$ is the square root of exactly one element (namely $g^2$)? $\endgroup$ Oct 21, 2015 at 11:51
  • $\begingroup$ I suppose you could look at it that way. $\endgroup$ Oct 21, 2015 at 12:00

The answer to Question 1) is "No" (at least when "information from the character table" is understood as only the matrix containing the values in the character table). The following is an example of two groups with "the same" character table, but different numbers of squares. Define $$ G_1 := \langle a_1, b_1, c_1\mid a_1^2=b_1^4=c_1^8=1,\; b_1^{a_1}=b_1^{-1},\; c_1^{a_1} = c_1^3,\; c_1^{b_1} = c_1^{-1}\; \rangle $$ and $$ G_2 := \langle a_2, b_2, c_2\mid a_2^2=b_2^4=c_2^8=1,\; b_2^{a_2}=b_2^{-1},\; c_2^{a_2} = c_2^{-1},\; c_2^{b_2} = c_2^3 \; \rangle .$$ Both groups are a semidirect product of $D_8 \cong \langle a_i, b_i \rangle$ acting on $C_8\cong \langle c_i \rangle$, but the action is different. (These are $\texttt{SmallGroup(64,149)}$ and $\texttt{SmallGroup(64,150)}$ from GAP's Small Groups Library.)

Then $G_1$ has $5$ squares, namely $\langle c_1^2 \rangle \cup \{b_1^2\}$, but $G_2$ has $6$ squares, namely $\langle c_2^2 \rangle \cup\{b_2^2, b_2^2c_2^4 =(b_2c_2)^2\}$, as one sees by computing a "general square" $(a_i^k b_i^l c_i^m)^2$ in both groups.

Finally, one can check (even without GAP) that the bijection $G_1\to G_2$ sending $a_1^ib_1^jc_1^k$ to $a_2^i b_2^j c_2 ^k$ induces a bijection between $\DeclareMathOperator{\Irr}{Irr} \Irr(G_1)$ and $\Irr(G_2)$ (the sets of irreducible characters, viewed as functions on $G_i$): The bijection between the groups yields an isomorphism $G_1/\langle c_1^4 \rangle \cong G_2/ \langle c_2^4 \rangle$, which takes care of the characters with $c_i^4 $ in the kernel. In both groups, the remaining characters are two characters of degree $4$ which vanish outside the center $\mathbf{Z}(G_i)=\langle b_i^2, c_i^4 \rangle$, so that the bijection above maps these characters onto each other: Indeed, $G_i/\langle b_i \rangle $ is isomorphic to the holomorph of $C_8$ in both cases, which has a unique faithful irred. character of degree $4$. The groups $G_1/\langle b_1^2c_1^4\rangle$ and $G_2/ \langle b_2^2 c_2^4 \rangle$ are not isomorphic, but both have a unique character of degree $4$ and central type.
(The $\chi\in \Irr(G_1/ \langle b_1^2 c_1^4 \rangle)$ with $\chi(1)=4$ has Frobenius-Schur indicator $\nu(\chi)=-1$, but the $\chi\in \Irr(G_2/\langle b_2^2c_2^4\rangle)$ with $\chi(1)=4$ has $\nu(\chi)=+1$. Such a thing must happen somewhere, as can be seen from the answer of Geoff Robinson.)


This is a supplement to YCor's answer, more like an extended comment.

Here is a different argument that, for $n\geq 4$, the number of squares in $S_n$ is even. By pairing each square with its inverse, we can restrict to the squares whose square is the identity. These are precisely the elements which are the product of an even number of pairwise disjoint transpositions. So let $e_n$ (resp. $o_n$) denote the number of elements of $S_n$ that are the product of an even (resp. odd) number of pairwise disjoint transpositions. Then, it suffices to show that $e_n$ and $o_n$ are both even for $n\geq 4$. This is easy by induction based on the following recursion for $n\geq 3$: \begin{align*} e_n&=e_{n-1}+(n-1)o_{n-2}\\ o_n&=o_{n-1}+(n-1)e_{n-2}\end{align*} The recursion is straightforward to verify by distinguishing on whether a given element of $\{1,\dots,n\}$ participates in a transposition or not.

  • $\begingroup$ Yes it's indeed simpler. I also checked that for $n\ge 8$ the number of squares if divisible by 4, but the proof is quite complicated and possible generalizations (e.g. studying being divisible by 8 for $n\gg 1$ [$n\ge 8$ seems OK]) would be even more messy. It would be interesting if there is a simpler approach. $\endgroup$
    – YCor
    Oct 18, 2015 at 11:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .