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This is Miklos Schweitzer 2009 Problem 6. It's a group theory problem hidden in a complicated language.

A set system $(S,L)$ is called a Steiner triple system if $L \neq \emptyset$, any pair $x,y \in S, x \neq y$ of points lie on a unique line $\ell \in L$, and every line $\ell \in L$ contains exactly three points. Let $(S,L)$ be a Steiner triple system, and let us denote by $xy$ the third point on a line determined by the points $x \neq y$. Let $A$ be a group whose factor by its center $C(A)$ is of prime power order. Let $f,h:S \to A$ be maps, such that $C(A)$ contains the range of $f$, and the range of $h$ generates $A$. Show that if $$ f(x)=h(x)h(y)h(x)h(xy)$$ holds for all pairs of points $x \neq y$, then $A$ is commutative and there exists an element $k \in A$ such that $$ f(x)= k h(x),\ \forall x \in S $$

Here is what I've got:

  • Because the image of $h$ generates $A$, for $A$ to be commutative is enough to prove that $h(x)h(y)=h(y)h(x)$ for every $x,y \in S$.

  • For the last identity to be true (if we have proved the commutativity) it is enough to have that the product $h(x)h(y)h(xy)=k$ for every $x \neq y$.

  • $h(y)h(x)h(xy)=h(xy)h(x)h(y)$

  • I should use somewhere the fact that the factor $A /C(A)$ has prime power order.

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I think $A/C(A)$ having prime power order is superfluous. I think the main thing to remember is that all three points of a line are basically indistinguishable. Let me know if you want a solution written up. –  Jack Schmidt Jun 22 '11 at 20:19
I don't (necessarily) want a complete solution. Just a few steps towards the solution. –  Beni Bogosel Jun 22 '11 at 20:21
Definitely then just get into the zen of {x,y,xy} is a set of three indistinguishable points. (xy)(y) = x, (y)(xy) =x and so on. You can leverage what you've already done into a complete proof. I found it rewarding to define $h(\ell)$ to be the product of the values of $h$ on the points of the line, though I don't think it would need to be included in a final answer. –  Jack Schmidt Jun 22 '11 at 20:28
Ok. Thanks. I'll try that, –  Beni Bogosel Jun 22 '11 at 20:31
@Beni: possibly never mind. I looked back at my notes, and I think I misread my handwriting early on. I decided to rewrite it without $h(\ell)$, and things went there (badly) quickly. If h is well-defined on lines (even to A/C(A)), then everything is nice and smooth, but I'm not sure how to prove that now. –  Jack Schmidt Jun 22 '11 at 21:30

2 Answers 2

up vote 5 down vote accepted

Let $A^1 := A/C(A)$ and $h^1(x)$ the equivalence class in $A^1$ which $h(x)$ belongs to. For each pair $x, y\in S$ we have $$ \begin{gather} h^1(x)^2 = e \\ h^1(x)h^1(y)h^1(x)h^1(xy) = e \tag 1\\ h^1(xy)h^1(y)h^1(xy)h^1(x) = e \end{gather} $$ From the previous equalities we can conclude that if $h^1(x) \in C(A^1)$ then $$ h^1(x) = h^1(y) $$ for each $y\in S$, and so we can rewrite $h(x)$, for each $x\in S$, as $$ h(x) = c(x)k $$ where $k\in A$ and $c(x)\in C(A)$. Naturally we have: $$ h(x)h(y) = c(x) k c(y) k= c(y) k c(x) k = h(y)h(x) $$ therefore $h(x)\in C(A)$ for each $x\in S$ and the problem is solved.

Now we must care about the case when no $h^1(x)$ belongs to $C(A^1)$.

Let $A^2 := A^1/C(A^1)$ and $h^2(x)$ the equivalence class in $A^2$ which $h^1(x)$ belongs to.
Since $o(A^1) = 2^{n_1}$, $C(A^1) \ne \{ e \}$ and $o(A^2) = 2^{n_2}$ with $n_2 < n_1$.
Moreover for each pair $x, y\in S$, $h^2(x)$, $h^2(y)$ satisfy relations similar to (1).

At this point, we can repeat for $A^2$ and $h^2(x)$ the previous reasoning and conclude that if $h^2(x)\in C(A^2)$ for some $x\in S$ then $h^1(x)\in C(A^1)$, contrary to the assumption.
Thus we must suppose no $h^2(x)$ belongs to $C(A^2)$.
Then we construct the group $A^3 := A^2/C(A^2)$ with order $o(A^3) = 2^{n_3}$ and $n_3 < n_2$, and repeat once more the previous steps.

Iterating the above procedure we will arrive to an abelian group $A^i$ (if not before, when $o(A^i)\le 4$) and to a contradiction so the commutativity of A will be established.

Being $A$ commutative, for each pair $x,y\in S$ we can write: $$ g(x) = h(xy) g(y) h(xy)^{-1} = g(y) $$ where $g(x) := h(x)^{-1}f(x)$, and therefore for each $x\in S$ $$ f(x) = k h(x) $$

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What theorem is the basis of your claim that $C(A^1)\neq \{e\}$? –  Thomas Andrews Jun 26 '11 at 19:02
@Thomas: It's a standard result that a group of prime power order has non trivial center. See, for example, the following link: Abstract algebra - W. E. Deskins –  AlbertH Jun 26 '11 at 19:46
Thanks, I had forgotten that result. –  Thomas Andrews Jun 26 '11 at 19:57
@Thomas: You are welcome... If you knew, how many results I forget... –  AlbertH Jun 26 '11 at 20:32

Let $g:S\rightarrow A$ be defined as $g(x) = h(x)^{-1} f(x)$.

Now, if $\{x,y,z\}\in L$, then $g(y) = h(z)g(x)h(z)^{-1}$. This means that the image of $g$ is closed under conjugation by elements of $A$ since $A$ is generated by the image of $h.$

Also, since this formula does not depend on the order of $x,y,z$, it means that $g(x)=h(z)g(y)h(z)^{-1}$. In particular, then $h(z)^2$ commutes with $g(x)$ for all $x$.

But since $f(x)$ is in the center of $A$, that means that $h(z)^2$ commutes with $h(x)$ for all $x, z\in S$. Hence $h(z)^2$ commutes with all of $A$ - that is $h(z)^2\in C(A)$, so $A/C(A)$ is generated by elements of order $2$, so by the condition of the problem, $A/C(A)$ must be of order $2^n$ for some $n$.

Now, since $g(x)=h(y)h(x)h(z) = h(z)h(x)h(y)$, we can see that:

$$g(x)^2 = h(y)h(x)h(z)h(z)h(x)h(y) = h(x)^2 h(y)^2 h(z)^2$$

Therefore, $g(x)^2 = g(y)^2 = g(z)^2$, and in particular, for all $x,y \in S$, $g(x)^2 = g(y)^2$. So there is some $K\in C(A)$ such that $\forall x\in S, g(x)^2=K$.

There are lots of things that can be concluded from knowing that $h(x)^2\in C(A)$. For example, that $f(x)f(y) {f(z)}^{-1}= h(x)^2 h(y)^2$. That can be used to show that $f(x)f(y)f(z) = h(x)^4h(y)^4h(z)^4 = K^2$.

Not sure where to go from here.

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