A semigroup $S$ is defined to be squared if there exists a subset $A\subseteq S$ such that the function $A\times A\to S$, $(x,y)\mapsto xy$, is bijective.

Problem: Is each squared finite group trivial?

Remarks (corrected in an Edit).

  1. I learned this problem from my former Ph.D. student Volodymyr Gavrylkiv.

  2. It can be shown that a group with two generators $a,b$ and relation $a^2=1$ is squared. So the adjective finite is essential in the above problem.

  3. Computer calculations show that no group of order $<64$ is squared.

  4. For any set $X$ the rectangular semigroup $S=X\times X$ endowed with the binary operation $(x,y)*(a,b)=(x,b)$ is squared. This follows from the observation that for the diagonal $D=\{(x,x):x\in X\}$ of $X\times X$, the map $D\times D\to S$, $(x,y)\mapsto xy$, is bijective. So restriction to groups in the formulation of the Problem is also essential.

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    $\begingroup$ I don't see how the free group on two generators can be squared. If $A$ contains the identity, then injectivity is contradicted by $1a=a1$ for $1\neq a\in A$. Surjectivity implies that $A$ contains some element $a$ and its inverse. But then injectivity and $aa^{-1}=a^{-1}a$ implies that $a=a^{-1}$. But a free group contains no elements of order $2$. $\endgroup$ Commented Feb 18, 2021 at 9:11
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    $\begingroup$ @JeremyRickard Of course you are right concerning the free group with two generators. The initial problem was for finite groups. I just wanted to present some simple example of a squared infinie groups and not thinking too much have written the free group. I hope that an infinite squared group can be constructed as the group with two generators $a,b$ and a relation $w^2=1$ for a suitable word $w\in A$, maybe even $w=a$. Now I will think a bit more in order to be sure. $\endgroup$ Commented Feb 18, 2021 at 9:42
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    $\begingroup$ Would you mind saying a word about how the infinite example is squared? i.e. what is $A$ in that case? $\endgroup$ Commented Feb 19, 2021 at 13:45
  • $\begingroup$ Just a somewhat trivial but maybe helpful observation: In the finite groups case, the condition is equivalent to: "There exists $A\subseteq G$ with $|G|=|A|^2$ such that, in the Cayley graph $Cay(G,A)$, any two vertices are connected by a directed path of length 2." (If you're hunting for a non-trivial example, this formulation feels easier to deal with.) $\endgroup$ Commented Feb 19, 2021 at 14:03
  • $\begingroup$ @AchimKrause The set $A$ includes $a$ and the sequence of words $w_n$ and $w_n^{-1}g_{m_n}$ where $\{g_m\}_{m\in\omega}$ is an enumeration of the group and $(w_n)$ is a sequence of long words chosen by induction in order to make the map $A\times A\to G$ bijective. I hope that this should work (but have not written the detail proof yet). $\endgroup$ Commented Feb 19, 2021 at 14:06

3 Answers 3


I think, as implicitly suggested by Yemon Choi, it is possible to explain the proof of the answer of user49822 by making more use of idempotents. Suppose that the finite group $G$ is squared via the subset $A$. The element $ e = \frac{1}{|G|}\sum_{g \in G} g$ is a primitive idempotent of $\mathbb{C}G.$

Let $ f= \frac{1}{|A|}\sum_{a \in A} a.$ Then we have $f^{2} = ef = fe = e = e^{2}$. Thus $(f-e)^{2} = 0 = e(f-e) = (f-e)e.$ Now $f = e +(f-e)$ is the sum of commuting matrices (in the regular representation of $\mathbb{C}G$, say) with the second matrix nilpotent.

Thus $f$ has trace $1$ in the regular representation of $\mathbb{C}G.$ This forces $1 \in A$ since all non-identitiy elements of $G$ have trace zero in the regular representation. But then $A = \{1 \}$, since (as in Jeremy Rickard's comment) if $a \neq 1 \in A$, then $a = 1a = a1$ gives two different expressions for $a$. Alternatively, (using traces) the fact that $1$ appears with coefficient $|G|^{-1}$ in $e$ tells us that $1$ appears with multiplicity $|G|^{-1}$ in $f$ as well, so that $\sqrt{|G|} = |A| = |G|$ and $|G| = 1.$

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    $\begingroup$ Why is $f = f^2$ ? $\endgroup$ Commented Feb 19, 2021 at 22:05
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    $\begingroup$ I think user49822's argument with characters can be interpreted as showing $fe' = 0$ when $e'$ is the primitive idempotent corresponding to any nontrivial irrep. Writing the identity as a sum of primitive idempotents gives $f = fe = e$. $\endgroup$
    – lambda
    Commented Feb 19, 2021 at 22:49
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    $\begingroup$ @NoamD.Elkies : Thanks. I amended the argument, but it is now less direct than I would have liked. $\endgroup$ Commented Feb 19, 2021 at 23:44
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    $\begingroup$ @GeoffRobinson Do you indeed need that $e(f-e)=(f-e)e$ in your argument? $\endgroup$ Commented Feb 20, 2021 at 6:22
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    $\begingroup$ Comment by Ville Salo in the other answer suggests that $2$ does not play a special role here. Let $A^n = A \times \cdots \times A$ ($n$ times, $n > 1$). If the multiplication map $A^n \rightarrow G$ is bijective, then $f^n = e$. Then $(f-e)^n = f^n - e = 0$, so $f-e$ is nilpotent and thus $f$ has trace $1$ in the regular representation. This implies $1 \in A$, so $A = \{1\}$ and $G = \{1\}$. $\endgroup$
    – spin
    Commented Feb 21, 2021 at 2:14

It seems that every squared finite group is indeed trivial.

Let $G$ be a squared finite group with the subset $A$ showing the squared-ness of $G$. For any irreducible representation $\pi$ of $G$, denote $u_\pi = \sum_{g\in A} \pi(g) \in \operatorname{End} V_\pi$ where $V_\pi$ is the vector space of the representation. Then, by the condition on $A$, we have $u_\pi ^2 = \sum_{x,y\in A}\pi(xy) = \sum_{g\in G}\pi(g)$, which is $0$ if $\pi$ is not trivial, and $|G|$ if $\pi$ is trivial. That is, if $\pi$ is not trivial then $u_\pi$ is nilpotent, hence $\operatorname{tr}u_\pi=0$.

Expanding $\operatorname{tr} u_\pi$, we get (for nontrivial $\pi$) $$ 0=\operatorname{tr}u_\pi=\sum_{g\in A}\chi_\pi(g) = \sum_{g\in G}\frac{\left|g^G\cap A\right|\chi_\pi(g)}{\left|g^G\right|} $$ where $g^G$ is the conjugacy class of $g$ in $G$, and the last equation follows from the fact that $\chi_\pi$ is a class function (constant on each conjugacy class). As the set of characters is an orthogonal basis of the space of class functions, it follows that the function $g\mapsto \frac{\left|g^G\cap A\right|}{\left|g^G\right|}$ is proportional to $\chi_1$, that is independent on $g$. In particular we can substitute $1$ in $g$ and get $$ \frac{\left|g^G\cap A\right|}{\left|g^G\right|} = \frac{\left|\left\{1\right\}\cap A\right|}{1} $$ Therefore,

  • if $1\in A$ then $g^G\subset A$ for any $g\in G$, which means that $A=G$.
  • On the other hand, if $1\not\in A$, then $g^G\cap A=\emptyset$ for any $g\in G$, which means that $A=\emptyset$.

As it is clear that $|A|^2=|G|$, it follows that the only possible case is $G=A=\left\{1\right\}$.

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    $\begingroup$ I guess that, like my argument (but for all groups and not just abelian ones), this really shows the stronger statement that if the multiplication map $A \times A \to G$ hits every element the same number of times then $A = \emptyset$ or $A = G$. $\endgroup$
    – lambda
    Commented Feb 19, 2021 at 16:52
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    $\begingroup$ I guess $2$ plays no special role here and no finite group has a root $\endgroup$
    – Ville Salo
    Commented Feb 19, 2021 at 17:02
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    $\begingroup$ I guess a restatement of this approach (and to some extent the one by @lambda ) is that the indicator function $1_A$ cannot be a convolution (square) root of the minimal idempotent in the complex group ring corresponding to the trivial rep? $\endgroup$
    – Yemon Choi
    Commented Feb 19, 2021 at 19:12
  • $\begingroup$ I think the answer would be clearer if you wrote e.g. $=\left\langle \chi_\pi, \frac{\lvert \cdot^G\cap A\rvert}{\lvert \cdot^G\rvert}\right\rangle$ in the equation after "expanding...". $\endgroup$
    – tomasz
    Commented Feb 21, 2021 at 18:11

Here is a proof for the abelian case, that perhaps has some chance to generalize.

Suppose $G$ is a squared finite group as witnessed by the subset $A$. Consider the element $$\alpha = \frac{1}{|A|} \sum_{a \in A} a$$ of the group algebra $\mathbb C G$. It is clear that $\alpha$ acts as the identity on the trivial representation. The squared condition implies that $$\alpha^2 = \frac{1}{|G|} \sum_{g \in G} g$$ which, as is standard, acts as the identity on the trivial rep and annihilates all other irreps. Thus $\alpha$ itself squares to zero on all nontrivial irreps.

In the abelian case where the irreps are 1-dimensional, this implies that $\alpha$ is just zero on the nontrivial irreps, so $\alpha = \alpha^2$ and $A = G$, which obviously only satisfies the bijectivity condition if $G$ is trivial. To extend this to the nonabelian case, one would have to rule out $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ Jordan blocks.

(Note that I didn't quite use the full power of the squared condition, only that $A \times A \to G$ is $|A|^2/|G|$-to-one. In particular this proves that the only subset with this property in finite abelian groups is $A = G$. But that stronger statement may well fail in the nonabelian case, so perhaps the fact that $|A|^2 = |G|$ needs to be used in some essential way.)

Edit: As was pointed out in a (now deleted) comment, the abelian case is in fact rather trivial since obviously commutativity already tells you that if $|A| > 1$ the map can't be injective. Of course proving the abelian case was not really the main reason for posting this, but I think it's worth mentioning anyway!

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    $\begingroup$ And while I was typing this answer someone did indeed come up with the general form of the same argument. Oh well. $\endgroup$
    – lambda
    Commented Feb 19, 2021 at 16:35

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