# Is each squared finite group trivial?

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.

• 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 $2$. Feb 18, 2021 at 9:11
• @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. Feb 18, 2021 at 9:42
• Would you mind saying a word about how the infinite example is squared? i.e. what is $A$ in that case? Feb 19, 2021 at 13:45
• 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.) Feb 19, 2021 at 14:03
• @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). Feb 19, 2021 at 14:06

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 $$|G| = 1.$$

• Why is $f = f^2$ ? Feb 19, 2021 at 22:05
• 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 $f = fe = e$. Feb 19, 2021 at 22:49
• @NoamD.Elkies : Thanks. I amended the argument, but it is now less direct than I would have liked. Feb 19, 2021 at 23:44
• @GeoffRobinson Do you indeed need that $e(f-e)=(f-e)e$ in your argument? Feb 20, 2021 at 6:22
• 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 $G = \{1\}$.
– spin
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 $$G=A=\left\{1\right\}$$.

• 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 $A = G$. Feb 19, 2021 at 16:52
• I guess $2$ plays no special role here and no finite group has a root Feb 19, 2021 at 17:02
• 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? Feb 19, 2021 at 19:12
• 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...". 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!

• And while I was typing this answer someone did indeed come up with the general form of the same argument. Oh well. Feb 19, 2021 at 16:35