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This question came up in teaching a course on basic group theory to high school students.

I gave the class the task of enumerating as many subgroups as they could find in $S_4$ and in the group $O$ of rotations of the cube. The students became suspicious that the groups were isomorphic, and came up with the idea of considering the rotations' actions on the cube's long diagonals to get a map $f:O\rightarrow S_4$, which they (rightly) suspected was an isomorphism.

The conversation stalled out when they tried to prove that it was an isomorphism. I directed them first to the question of whether $f(xy)=f(x)f(y)$, figuring that this would make it easier to deduce injectivity or surjectivity, e.g. by finding generators in the image. However, they were already stuck on whether $f(xy)=f(x)f(y)$, although when they actually calculated $f(xy)$ and $f(x)f(y)$ for two or three pairs of rotations $x,y$ (which involved some intense visualizing work to compose the rotations in 3-space, since they do not know about matrices), the answers did match. As I listened to them talk about it, I realized that there were some subtleties here having to do with the way we labeled the group elements that made me realize there was something fundamental I wanted to understand better here too.

I want to get to the bottom of this. Here's the setup:

We were looking at rotations of the cube in terms of a fixed coordinate system; you could see them as alibi transformations. Thus, for example, although we didn't do this, you could see the cube as the one with vertices $\{\pm 1,\pm 1,\pm 1\}$ in $\mathbb{R}^3$, and $O$ as the group generated by matrices

$$x=\begin{pmatrix} & &1\\1& & \\ &1& \end{pmatrix},\; y=\begin{pmatrix}1& & \\ & &-1\\ &1& \end{pmatrix}$$

The point is that we imagine that the coordinate system is fixed, the group elements are fixed alibi transformations in the coordinate system, and the cube is being moved around.

Meanwhile, we were writing down the permutations as functions from the index set $\{1,2,3,4\}$ to itself, and composing them accordingly, using the right-to-left convention of function composition. $(123)$ applied to $1$ equals $2$, etc.

The map $f$ was given by choosing some labeling on the long diagonals of the cube in its initial position, for example $\overline{(-1,-1,-1)(1,1,1)} = 1$, $\overline{(1,-1,1)(-1,1,-1)} = 2$, $\overline{(1,-1,-1)(-1,1,1)} = 3$, and $\overline{(1,1,-1)(-1,-1,1)} = 4$. Then for a given rotation $\rho$, defining $f(\rho)$ as the permutation that describes where $\rho$ sends each index when the cube is in its original position.

To illustrate, in the present example, this means $y\mapsto (1234)$, for example, and $x\mapsto (243)$. We have

$$xy = \begin{pmatrix} & &1\\1& & \\ &1& \end{pmatrix}\begin{pmatrix}1& & \\ & &-1\\ &1& \end{pmatrix} = \begin{pmatrix} &1& \\1& & \\ & &-1\end{pmatrix}$$

and $f(xy) = (14)$. Meanwhile, $f(x)f(y) = (243)(1234) = (14)$ (remember that we are composing permutations right to left).

Thinking about both the permutations and the rotations as functions, in the former case on the set of indices and in the latter case on the cube, and the action can be restricted to the long diagonals, it is clear (obvious) to me why $f$ is a homomorphism. All $f$ does is it records what a rotation $\rho$ does to the diagonals. So $f(\rho\eta)$ is what $\eta$, followed by $\rho$, does to the diagonals, and $f(\rho)f(\eta)$ is what $\eta$ does to the diagonals followed by what $\rho$ does to the diagonals. These are obviously the same.

But there is another way of thinking about it that I can engage in that confuses me about this, and I can't quite sort through where the problem is. Furthermore, this (clearly wrong) way of thinking deals with some issues that the above doesn't engage at all. So my question is:

(A) What is wrong with the following train of thought? (B) Can you explain in terms that engage with this train of thought why $f$ is a homomorphism?

As above, think of the group $O$ as acting on the cube by rotations in a fixed coordinate system. $\rho\in O$ refers to a rotation with reference to the fixed coordinate system; changes in orientation of the cube do not affect the meaning of $\rho$.

Choose an initial orientation of the cube, and a labeling of the long diagonals, as above. For each $\rho\in O$, write down the permutation $f(\rho)$ that describes the action of $\rho$ on the long diagonals in this initial orientation.

Now consider $f(\rho\eta)$. $\eta$ moves the diagonals somewhere, $\rho$ moves them again, and $f$ writes down what the composition did, in terms of the original orientation of the cube.

On the other hand, consider $f(\rho)f(\eta)$. $f(\eta)$ is unproblematic; it writes down what $\eta$ did to the diagonals, which is the same as before. But now the diagonals are in new spots, so when $\rho$ acts on them, it will not do the same thing it did when it acted on the cube in its original orientation. Thus what $\rho$ does to the diagonals after $\eta$ has been applied will not be the same as $f(\rho)$, which is trying to describe what $\rho$ did to the diagonals in their original orientation. Thus $f(\rho)f(\eta)$ should come out to something different.

Empirically, this logic is wrong, and furthermore it contradicts the above logic explaining why $f$ is (obviously) a homomorphism. I assume there is some kind of alias/alibi confusion in it. Be that as it may, I can't shake it completely. Can you help? An ideal answer would both explain where the fallacy is and also give a correct explanation of what's going on that engages with the alias/alibi issues in the fallacious logic.

Thanks in advance.

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    $\begingroup$ I suspect this is the same sort of confusion that happens when dealing with the "Fifteen puzzle" with numbered tiles. The tiles are numbered 1-15 (and a 16th empty space is blank, but maybe it is a good idea to label it 16). The moves involving moving adjacent tiles into the blank spot, so every move is a transposition of the form $(n,16)$ where $n$ is some tile next to the blank. -- How do you describe this in terms of a group, since the moves available change as the blank moves around? $\endgroup$ – Jack Schmidt May 16 '14 at 18:00
  • $\begingroup$ "But now the diagonals are in new spots, so when $\rho$ acts on them, it will not do the same thing it did when it acted on the cube in its original orientation. Thus what $\rho$ does to the diagonals after $\eta$ has been applied will not be the same as $f(\rho)$, which is trying to describe what $\rho$ did to the diagonals in their original orientation." .... This reads like gibberish to me, I have no idea what you're talking about. With $S_4$ acting on $\{1,2,3,4\}$, when considering what $\rho\sigma$ does, is it worrisome that nothing is in "its original position" after applying $\sigma$? $\endgroup$ – blue May 21 '14 at 2:43
  • $\begingroup$ I think a key point of evidence is missing. How exactly did you work it out "empirically"? $\endgroup$ – marshal craft Dec 4 '15 at 5:53
  • $\begingroup$ @marshalcraft - see the paragraph that begins "To illustrate..." The calculation in $O$ is done with standard matrix multiplication and the calculation in $S_4$ is done with standard permutation composition, adopting the right-to-left composition convention. $\endgroup$ – Ben Blum-Smith Dec 9 '15 at 5:24
  • $\begingroup$ If you have two linear transformations $A$ and $B$, then consider the composition of them $AB$, the inverse of the coordinate change is $B^{-1}A^{-1}$ but what is $A^{-1}B^{-1}$? $\endgroup$ – marshal craft Dec 10 '15 at 8:05
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Answer to (A) It is ambiguous and confusing to say that $f(\rho)$ “records what $\rho$ does to the diagonals” without distinguishing between diagonals as fixed positions in space and diagonals as movable objects in space, see below.

Answer to (B) Distinguish between a “geographical”, “unmovable”, “absolute” cube made of four “absolute” diagonals $AD_1,AD_2,AD_3,AD_4$ and a moving cube made of four “physical”, “movable” diagonals (we will call them rods) $MD_1,MD_2,MD_3,MD_4$. In every “reachable” position, each rod $MD_j$ has a unique location among the $AD_i$.

What does $f(\rho)$ express in this context ? If $f(\eta)$ sends $i$ to $j$, $\eta$ is to move the rod located in $AD_i$ (let us call it $r$) and put it at $AD_j$. In turn, if $f(\rho)$ sends $j$ to $k$, $\rho$ is to move again this same $r$ from $AD_j$ to $AD_k$. This succession of two moves, $\eta$ followed by $\rho$, is equivalent to simply move $r$ from $AD_i$ to $AD_k$. This should make it clear that $f(\rho\eta)=f(\rho)f(\eta)$.

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  • $\begingroup$ +1 This is exactly the kind of answer I was looking for. $\endgroup$ – Ben Blum-Smith Dec 9 '15 at 5:25
  • $\begingroup$ This answer hardly distinguishes the differences between alibi and alias, at least not in a useful way. You posted that the ideal answer should. Sorry if I come off overly "strong" I just would like to get at the bottom of the issue. It may be illuminating to do so. $\endgroup$ – marshal craft Dec 10 '15 at 8:16
  • $\begingroup$ @marshalcraft Personally I feel that my answer gets at the bottom of the issue and is illuminating. I don't mind your coming strong at all, I only mind what looks like mere showing off. Why don't you post your better answer $\endgroup$ – Ewan Delanoy Dec 10 '15 at 8:42
  • $\begingroup$ @marshalcraft - the answer engaged to my satisfaction with the train of thought in the question. The confusion I was having, I believe it turns out, was pretty much internal to viewing all transformations as alibi transformations and was not really an alias/alibi confusion at all. It stemmed from failing to fully distinguish between the locations and the things sitting in those locations, as this answer carefully makes clear. So the answer resolved my question. $\endgroup$ – Ben Blum-Smith Dec 11 '15 at 15:04
  • $\begingroup$ @marshalcraft - To your point, though: the confusion that Ewan Delanoy cleared up for me does remind me of the alias/alibi issue, and in a way that I don't feel I completely have a beat on. So it would indeed be illuminating to try to articulate what the substance of this similarity is. If you have thoughts about this I'd love to hear them. $\endgroup$ – Ben Blum-Smith Dec 11 '15 at 15:08

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