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I'm looking for a geometrical interpretation of the action of automorphisms of cyclic groups. I'll take one particular example to make it clear :

I'm taking the cyclic group $\mathbb{Z}_{12}$, which has $\mathbb{Z}_2 \times \mathbb{Z}_2$ as it automorphism group. Considering the integers modulo 12, the automorphisms are the multiplications by 1, 5, 7 or 11.

I'm considering now the action of $\mathbb{Z}_{12}$ on a discretized circle of 12 points in a plane (an dodecagon). The action of multiplying by 11 (left figure below) corresponds to a geometrical reflection in the plane, and I recover the dihedral group of symmetries of the dodecagon.

enter image description here

However, multiplication by 5 or 7 sends shapes drawn on this circle to different shapes, and I find no obvious geometrical meaning to these. Is there any ? (For example in the right figure above, the action of multiplying by 5 on the purple triangle gives the green triangle, with no obvious relation between them) Since these automorphisms can be considered as affine transformations, and affine spaces can be considered as subspaces of projective spaces, I'm considering working in the projective line over the field of 13 elements but for now I'm getting nowhere...


This post was edited as my original message was dealing with $\mathbb{Z}_8$. Interestingly, although $\mathbb{Z}_8$ has the same automorphism group, all multiplications correspond to geometrical reflections in the plane....

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I didn't quite understand what kind of a geometric object you are looking for in the general case. To get the octagon out of $\mathbf{Z}_8$ you label the vertices of the octagon with the elements of the group. Then you join two vertices, iff their labels differ by the chosen generator 1 of your group. Then you hope that if you apply an automorphism of the group to the labels of the vertices, you also get an automorphism of this graph? But for that to happen you also need to join the pairs of vertices with labels differing from each other by 3. Is that what you want? – Jyrki Lahtonen Aug 23 '11 at 19:05
I just edited my post, and sorry for not giving enough details, but yes, the dodecagon is labelled by powers of the element of order 12... As said in my question, I'd like to know if there is a geometrical interpretation to the distorsion of the green triangle (or any other for that matter) in the case of multiplication by 5 or 7... – AlexPof Aug 23 '11 at 20:37
The only reason you see geometry behind some some automorphisms and not others is because of the arbitrariness of the choice of generator. Change the generator, and you change which automorphims have an obvious geometric meaning. You could also up the number of dimensions you're working in until there exists a configuration where every automorphism identifies with an affine transformation (maximum dimension needed would be the order of the group, naturally). – anon Aug 23 '11 at 21:50
I'm still somewhat confused by what you mean, when you say that "I recover the dihedral symmetries of the dodecahedron''. I don't see the rotations of the dodecahedron arising as automorphism of the group $G=\mathbf{Z}_{12}$! Any automorphism of $G$ will keep the vertex labelled $e$ fixed. I do realize that the action of the group $G$ on the set of 12 vertices gives you the rotations. But you seem to want two groups, $G$ and $Aut(G)$, to both act on your graph in a compatible way in the sense that action by $G$ labels the vertices with elements of $G$, and then $Aut(G)$ acts on the labels?? – Jyrki Lahtonen Aug 24 '11 at 5:59
@anon: I think I see what you're saying (which is similar to the answers below). If I understand correctly, I should find a representation $\rho$ of $\mathbb{Z}_{12}$ in a vector space V, such that there exist a single $M_{\psi}$ $\in$ GL(V) such that $\forall g \in \mathbb{Z}_{12}, \rho(\psi(g))=M_{\psi}*\rho(g)$ (with $\psi$ an automorphism) ? Is there a construction or brute force should be applied :) ? – AlexPof Aug 24 '11 at 13:00
up vote 2 down vote accepted

You can see the automorphisms of $Z_{12}$ as nice geometric transformations if you are willing to treat them separately.

-The identity (multiplication by 1) is always "geometric".

-Multiplication by -1 (or 11) is geometric in that it is realized by a reflection in the dodecahedron in your original post: the reflection fixes 0 and 6 = -6 (mod 12).

-Multiplication by 5 can be realized as follows. First count the fixed points. In other words solve $5x = x$ or $4x = 0$ in $\mathbb{Z}_12$. The solutions are 0, 3, 6, and 9. Now, view the group as the vertices of a rectangular hexagonal prism. Let $P$ be the prism, and let $H_b$ and $H_t$ be the bottom and top hexagons. Label the bottom vertices 0, 2, 4, 6, 8, 10 (cyclically) and label the top vertices 3, 5, 7, 9, 11, 1 so that 0 is joined to 3 by a vertical edge of the prism, 2 is joined to 5, etc. The underlying graph is almost the Cayley graph of $Z_{12}$ with respect to the (additive) generating set $\{2,3\}$; add the additional edges for the generator 3 to complete the Cayley graph. Multiplication by 5 is realized by a reflection. The fixed points are 0, 3, 6, and 9.

-Multiplication by 7 is realized in the same fashion. The fixed points are the solutions to $6x = 0$, i.e. 0, 2, 4, 6, 8, and 10. This time consider the Cayley graph with respect to the generating set $\{3, 4\}$. There is an nice reflection of this graph with the desired fixed points. (You should see three squares and 4 triangles; it can be nicely visualized on a torus, for instance.)

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Thanks, this answer really speaks to me, but I like Vladimir's too... – AlexPof Aug 25 '11 at 7:10

You can't say that you are considering THE action of $\mathbb Z_8$ on the circle because there are FOUR such faithful actions and three unfaithful actions. For the faithful ones, the generator, 1, can either be rotation by $\frac{2\pi}8=45^\circ$, rotation by $3\cdot\frac{2\pi}8=135^\circ$, rotation by $5\cdot\frac{2\pi}8=225^\circ$, or rotation by $7\cdot\frac{2\pi}8=315^\circ$. For the unfaithful ones, it can be rotation by $90^\circ$, $270^\circ$ or $180^\circ$.

The first geometric interpretation of the four automorphisms of $\mathbb Z_8$ is that they allow you to pass from one choice of faithful action to another choice of faithful action. In general, the endomorphisms of $\mathbb Z_8$ (which are given by the monoid of $\mathbb Z_8$ considered multiplicatively) allow one to pass from one choice of action to another choice of action. This probably does not seem terribly satisfactory, so below is a second geometric interpretation.

Each of the endomorphisms of $\mathbb Z_8$ corresponds (irrelevant of faithful action) to an endomorphism of the circle group, and endomorphisms of the circle group can be thought of as winding the plane around itself $x$ times.

So if you consider the set of vertices of your octagon, the geometric interpretation of an endomorphism $x\in\mathbb Z_8$ is that it acts on the vertices by winding the plane around itself $x$ times. This is why you the octagon becomes a star, one is the winding of the other.

The reason why the octagon is not preserved (but its set of vertices is) is that winding around actually changes the angles (e.g. $45^\circ$ become $3\cdot45^\circ=135^\circ$) and so distances must also change.

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I voted for the other answer, but I like this answer too. If I could have voted for both, I would have... – AlexPof Aug 25 '11 at 7:09
Thanks, AlexPof. There is a difference between voting for and accepting an answer though: voting is clicking on the up-arrow and indicates how useful the answer is, accepting (clicking the checkmark) is whether that answer is the one you're looking for. Though you can accept only one answer to a question you ask, you can vote for as many answers as you like (apologies if you already know this and have simply chosen not to use the voting mechanism; your profile seems to indicate you haven't properly voted for anything). – Vladimir Sotirov Aug 25 '11 at 13:23
Sorry, I meant accepting... and I didn't think of accepting answers myself, I thought it was more or less a ranking system by other members... – AlexPof Aug 26 '11 at 19:06
Coming back to another of my issues, am I really on the wrong way when trying to find a relation with projective spaces ? For example,… considers the holomorph of a cyclic group of power prime order as an affine group of transformations over the corresponding field. However, 12 is not a prime power, so that's why I was considering working with the projective space over $\mathbb{F}_{11}$ or $\mathbb{F}_{13}$ – AlexPof Aug 26 '11 at 19:35

An automorphism for a cyclic group needs to take a generator onto a generator.
An automorphism is also part of a symmetry group which for the bounded plane, can only be the identity, a rotation, or a reflection.

By applying these definition, we obtain a straightforward geometric visualization of $Aut[\mathbb Z_{12}]$.

Start with your $t^1$ generator.

  • Our first symmetry is identity or mapping everyone onto itself. This is the trivial $1$-automorph which takes $t^1$ to $t^1$. Alternatively, this can be seen as a $360^o$ rotation.

  • Now, fix a reflection line from $t^0$ to $t^6$. This reflects $t^1$ unto $t^{11}$. Hence the $11$-automorph corresponds to a reflection along the vertical line.

  • Next, fix a reflection line from $t^3$ to $t^9$. This reflects $t^1$ unto $t^5$. Hence the $5$-automorph corresponds to a reflection along the horizontal line.

  • Finally, rotate by $180^o$. This takes $t^1$ unto $t^7$. One can alternatively view this as a compound transformation by first reflecting on the vertical (horizontal) and then reflecting on the horizontal (vertical). Either way, this corresponds to the $7$-automorph.

To see this more clearly, map out the generators on your clockface and connect that dots. You will have a rectangle joining $t^1$, $t^5$, $t^7$, and $t^{11}$. Now you can clearly see that the v-reflection, h-reflection, $\pi$-rotation, and $2\pi$-rotation are the only transformations that preserve the rectangle, the only ways to map generators onto generators.

$11$-automorph defines inverses

Note that that the v-reflection, the 11-automorph, defines the inverses of the elements of the group $\mathbb Z_{12} $:

\begin{matrix} t^{11} = t^{-1}\\ t^7 = t^{-5} \\ t^{11} \not= t^{-5} \not= t^{-7} \\ t^{7} \not= t^{-1} \not= t^{-11}\\ \end{matrix}

We can verify this by considering the automorphisms of the $\mathbb Z_{12}$ subgroups, which should be invariant under the parent group automorphism.

Subgroups $\mathbb Z_{1}$ and $\mathbb Z_{2}$ have trivial $Aut[\mathbb Z_1]$.

Subgroups $\mathbb Z_3, \mathbb Z_4,$ and $\mathbb Z_6$ have non-trivial $Aut[\mathbb Z_2]$.

Now map $Aut[\mathbb Z_{12}]$ onto the non-trivial subgroups and notice that you will always see the $1$-automorph opposite the $11$-automorph and the $5$-automorph opposite the $7$- automorph. You will never see the $1$- and $11$- or the $5$- and $7$- automorphs mapped onto the same the subgroup automorph.

This mapping also gives insight into what it means to be have trivial $Aut[\mathbb Z_1]$. Do the mapping procedure on subgroups $\mathbb Z_1$ and $\mathbb Z_2$ and you'll note that in both cases, $Aut[\mathbb Z_{12}]$ maps to the same generator.

Additive and multiplicative interpretation of $Aut[\mathbb Z_{12}]$

Another observation is that interpreting $Aut[\mathbb Z ]$ as an additive group gives you the generators while interpreting it as a multiplicative group gives you the permutation symmetries. I'm not sure about the wording of the latter but as an example, in the case of $\mathbb Z_{12} $ the multiplicative interpretation means that the ${1,5,7,}$ and $11$-automorphs correspond to:

\begin{matrix} x \rightarrow x \\ x \rightarrow 5x \\ x \rightarrow 7x \\ x \rightarrow 11x \\ \end{matrix}

Finally, all of this is a refection (pun intended) of the fact that
\begin{matrix} Aut[\mathbb Z_{12} ] = Aut[\mathbb Z_{8} ]= \mathbb Z_2 \otimes \mathbb Z_2 \end{matrix}
which is the Klein 4-Group with two reflections ($\mathbb Z_2$) plus two rotations ($\mathbb Z_2$)

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