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Famously from every finite projective plane $P$ we can create a dual projective plane $P'$ by taking the points of $P'$ to be the lines of $P$ and the lines of $P'$ to be the points of $P$.

It is also well known that the smallest projective plane, the Fano plane, is isomorphic to its own dual. I am interested in the set of all isomorphisms between the Fano plane and its dual.

Perhaps an easier way to think about these maps is to take the Heawood graph: the bipartite graph whose 14 vertices represent the 7 points and 7 lines of the Fano plane with an edge connecting a point-vertex and a line-vertex whenever in the Fanoplane the point lies on the line.

This graph has 336 graph-automorphisms: 168 mapping point-vertices to point-vertices and line-vertices to line-vertices - a normal subgroup of the automorphsim group isomorphic to the symmetry group of the fano plane. And it has 168 automorphisms mapping point-vertices to line-vertices and vice-versa: the unique non-group coset for this normal subgroup. It are these 168 symmetries in the non-trivial coset that I'm interested in. My question is:

Are these all of order 2?

Obviously they are all of even order, but all being of order 2 is a much stronger claim. I don't see a clear reason for this when merely staring at the Fano plane, but in the mean time I wasn't able to construct an example of higher order. I hope someone can clarify this!

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    $\begingroup$ No -- Isomorphisms are in bijection to the automorphism group which has e.g. also elements of order 3 and 7. $\endgroup$
    – ahulpke
    Jan 17 at 22:38
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    $\begingroup$ It is a bijection, but why would the bijection respect the order? As a loose analogy: in the symmetries of the $n$-gon there is a bijection between the $n$ rotations and the $n$ reflections, but that doesn't mean relections have order other than 2 $\endgroup$
    – Vincent
    Jan 17 at 22:39

2 Answers 2

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Let $G$ be the full automorphism group of the Heawood graph and $H$ the subgroup induced from the automorphisms of the Fano plane. Let $\tau$ be a representative of a nontrivial coset.

Suppose all elements outside of $H$ have order two. Then $e=(\tau h)^2=\tau h\tau h=(\tau h\tau^{-1})h$ implies $\tau h\tau^{-1}=h^{-1}$ for all $h\in H$ (in analogy with the dihedral symmetries of a regular polygon, as you note in the comments). But this implies $h\mapsto h^{-1}$ is an automorphism of $H$, which is equivalent to $H$ being abelian. But $H\cong\mathrm{PGL}_3\mathbb{F}_2$ is not abelian, a contradiction.

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  • $\begingroup$ Ah this is great. $\endgroup$
    – Vincent
    Jan 17 at 22:48
  • $\begingroup$ Still something is bugging me here, but I can't quite put my finger on it. I'll come back to it later $\endgroup$
    – Vincent
    Jan 17 at 22:49
  • $\begingroup$ No this is obviously correct. Thanks for the quick answer! $\endgroup$
    – Vincent
    Jan 17 at 22:57
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Here is an attempt to turn Runway44's excellent answer into a concrete counterexample. We'll make use of the 'exceptional isomorphism' between $\mathbb{F}_2^3$ and $\mathbb{F}_8$ to find a $h \in H$ and $\tau \in G, \tau \notin H$. Our $\tau$ will have order 2, and hence not be a counterexample. However, following Runway44's answer we know that $\tau h$ will be our counterexample unless it equals $h^{-1}\tau$. And in that case runway's answer tells us that $\tau h'$ will be a counterexample for any $h' \in H$ not commuting with $h$. So let's go.

Since $\mathbb{F}_8$ is a field, the non-zero elements form a cyclic group under multiplication. Let $\alpha$ be a generator. It is known that it satifies $1 + \alpha + \alpha^3 = 0$. It follows that every non-zero element of $\mathbb{F}_8$ is of the form $\alpha^k$ and that $\alpha^k + \alpha^{k+1} + \alpha^{k+3} = 0$ where the $k$'s are read modulo 7.

By definition the points of the Fano plane are lines through the origin in $\mathbb{F}_2^3$ and the lines of the Fano plane are planes through the origin in $\mathbb{F}_2^3$. Since 2 is a very small number we can identify the set of points of the Fano plane with the non-zero elements of $\mathbb{F}_2^3$ and each individual Fano line with the three non-zero points in the corresponding plane. Such triples can be recognized by the fact that non-zero elements $a, b, c$ lie in the same plane if and only if $a + b + c = 0$.

Combining this with the beginning of the post we find that we can identify the points of the Fano-plane with the elements $\alpha^k$ for $k = 0, \ldots, 6$ and the lines with the triples $\{\alpha^k , \alpha^{k+1}, \alpha^{k+3}\}$ with $k$'s, again, read mod 7. This immediately suggests a candidate for an automorophism $h \in H$ of the Fano-plane:

$$h: \alpha^k \mapsto \alpha^{k+1}$$

Now in order to create an example for $\tau$ we switch back to the $\mathbb{F}_2^3$ picture. $\{1, \alpha, \alpha^2\}$ is a basis for this vector space. We let $\tau$ be the map that sends each line or plane through the origin to its ortho-complement under the 'standard inner product' with respect to this basis. The standard inner product is not technically an inner product in this case, but it is still a non-degenerate biliear form so the notion of orthocomplement is well-defined. E.g. the line through $(0, 0, 0)$ and $(0, 1, 0)$ is orthogonal to the plane $\{(0, 0, 0), (1, 0, 0), (0, 0, 1), (1, 0, 1)\}$ which in the $\mathbb{F}_8$ picture of the Fano-plane means that

$$\tau(\alpha) = \{1, \alpha^2, 1 + \alpha^2\} = \{\alpha^6, \alpha^0, \alpha^2\}$$

Similarly we have:

$$\tau(1) = \{\alpha, \alpha^2, \alpha + \alpha^2\} = \{\alpha^1, \alpha^2, \alpha^4\}$$

So far so good. Let's compute $(\tau h)(1)$ and $h^{-1}\tau(1)$ and see if they are the same or different:

$$(\tau h)(1) = \tau(\alpha) = \{\alpha^6, \alpha^0, \alpha^2\}$$

and

$$(h^{-1}\tau)(1) = h^{-1}(\{\alpha^1, \alpha^2, \alpha^4\}) = \{\alpha^0, \alpha^1, \alpha^3\}$$

So indeed they are different and hence $\tau h$ is not of order 2!

Computing what is the order of $\tau h$ is a bit more work, maybe I'll do that later.

UPDATE: I did the computation and the order of $\tau h$ equals 8.

COMPUTATION:

I'll start with a more modest goal and compute the length of the orbit of the element 1 under iterations of $\tau h$, so that at least we obtain a divisor of the order of $\tau h$.

We already established that $(\tau h)(1) = \{\alpha^6, \alpha^0, \alpha^2\}$. It follows that:

$$(\tau h)^2(1) = (\tau (h(\{\alpha^6, \alpha^0, \alpha^2\}))) = \tau(\{\alpha^0, \alpha^1, \alpha^3\})$$

Switching back to vector notation we get:

$$(\tau h)^2(1) = \tau(\{(1, 0, 0), (0, 1, 0), (1, 1, 0)\}) = (0, 0, 1) = \alpha^2.$$

We see that $(\tau h)^2(1) = h^2(1)$. Could it be the case that $(\tau h)^2 = h^2$ in general? Alas not: in that case we would have $\tau h = h \tau$ and evaluating both sides at 1 yields different answers. We conclude that the length of the orbit is at least 4...

$$(\tau h)^3(1) = (\tau (h(\alpha^2))) = \tau(\alpha^3)$$

Switching back to vectors we find

$$(\tau h)^3(1) = \tau((1, 1, 0)) = \{(0, 0, 1), (1, 1, 0), (1, 1, 1))\} = \{\alpha^2, \alpha^3, \alpha^5\}$$

It follows that

$$(\tau h)^4(1) = \tau(\{\alpha^3, \alpha^4, \alpha^6\} = \tau(\{(1, 1, 0), (0, 1, 1), (1, 0, 1)\}) = (1, 1, 1) = \alpha^5$$

This is getting annoying, I hoped I would be done by now. Anyway...

$$(\tau h)^5(1) = (\tau (h(\alpha^5))) = \tau(\alpha^6)$$ $$ = \tau((1, 0, 1)) = \{(1, 1, 1), (1, 0, 1), (0, 1, 0)\} = \{\alpha^5, \alpha^6, \alpha^1\}$$

so that

$$(\tau h)^6(1) = \tau(\{\alpha^6, \alpha^0, \alpha^2\}) = \tau(\{(1, 0, 1), (1, 0, 0), (0, 0, 1)\}) = (0, 1, 0) = \alpha$$

and

$$(\tau h)^7(1) = (\tau (h(\alpha))) = \tau(\alpha^2)$$ $$ = \tau((0, 0, 1)) = \{(1, 0, 0), (0, 1, 0), (1, 1, 0)\} = \{\alpha^0, \alpha^1, \alpha^3\}$$

so that

$$(\tau h)^8(1) = \tau(\{\alpha^1, \alpha^2, \alpha^4\}) = \tau(\{(0, 1, 0), (0, 0, 1), (0, 1, 1)\}) = (1, 0, 0) = 1!$$

So the length of the orbit is 8 and the order of $(\tau h)$ is some multiple of 8 that is still a divisor of 336.

But actually we got a bit more out of this computation than we asked for. $(\tau h)^2$ is an automorphism of the Fano plane whose order is a multiple of 4. But automorphisms of the Fano plane can only have orders 1, 2, 3, 4 or 7 as Wikipedia is kind enough to tell us. So in fact $(\tau h)^2$ has order exactly 4 and hence $\tau h$ has order 8!

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