48
$\begingroup$

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.

$\endgroup$
5
  • 8
    $\begingroup$ 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 $2$. $\endgroup$ – Jeremy Rickard Feb 18 at 9:11
  • 2
    $\begingroup$ @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. $\endgroup$ – Taras Banakh Feb 18 at 9:42
  • 2
    $\begingroup$ Would you mind saying a word about how the infinite example is squared? i.e. what is $A$ in that case? $\endgroup$ – Achim Krause Feb 19 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.) $\endgroup$ – Matt Zaremsky Feb 19 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). $\endgroup$ – Taras Banakh Feb 19 at 14:06
24
$\begingroup$

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 $|G| = 1.$

$\endgroup$
7
  • 4
    $\begingroup$ Why is $f = f^2$ ? $\endgroup$ – Noam D. Elkies Feb 19 at 22:05
  • 2
    $\begingroup$ 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 Feb 19 at 22:49
  • 1
    $\begingroup$ @NoamD.Elkies : Thanks. I amended the argument, but it is now less direct than I would have liked. $\endgroup$ – Geoff Robinson Feb 19 at 23:44
  • 1
    $\begingroup$ @GeoffRobinson Do you indeed need that $e(f-e)=(f-e)e$ in your argument? $\endgroup$ – Taras Banakh Feb 20 at 6:22
  • 3
    $\begingroup$ 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 Feb 21 at 2:14
43
$\begingroup$

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 $G=A=\left\{1\right\}$.

$\endgroup$
4
  • 6
    $\begingroup$ 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 Feb 19 at 16:52
  • 5
    $\begingroup$ I guess $2$ plays no special role here and no finite group has a root $\endgroup$ – Ville Salo Feb 19 at 17:02
  • 1
    $\begingroup$ 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 Feb 19 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 Feb 21 at 18:11
14
$\begingroup$

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!

$\endgroup$
1
  • 6
    $\begingroup$ And while I was typing this answer someone did indeed come up with the general form of the same argument. Oh well. $\endgroup$ – lambda Feb 19 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.