# Inner Automorphisms and Outer Automorphisms

An inner automorphism is an automorpism of the form p(g)=G^h=h'gh for all g in G and for a specific, fixed h in G. An outer automorphism is an automorphism that is not inner.

But occasionally the definition takes a slightly different form. The alternate definition says that automorphisms are of the form p(g)=G^h=h'gh for all g in G. If h is a fixed element of G, then p is an inner automorphism. Otherwise, h is not in G but rather is a fixed element of a larger group of which G is a subgroup, and p is an outer automormphism. The latter definition clearly motivates the names "inner" and "outer".

Are the two definitions equivalent?

I was reminded of this question because of the recent discussion of distance preserving automorphisms of the cube. The distance preserving automorphism G^m, where m is a symmetry of the cube, is an outer automorphism because m is not in G (except in the trivial case where m is the identity permutation). But if we consider the larger group G

_{M}that includes G and all its symmetries, then G

_{M}^m is an inner automorphism because m is in G

_{M}by definition.

Jerry

## Comment viewing options

### There are a number of ways to

_{M}, so there is no doubt about its existence. It seems to me that the most straightforward way to construct it is as follows.

First, if Q is the set of quarter turns, then we may simply generate G as G=<Q>.

Second, we need generators for M, the group of 48 cube symmetries. Unfortunately, Cube-Lovers never settled on a standard notation for the individual elements of M, so let's try the following. We may think of M as C ∪ Cν, where C is the group of 24 cube rotations, and ν is the central inversion. So Cν is the set (not group) of 24 reflections of the cube. We let c

_{r}be a 90 degree clockwise rotation of the whole cube centered on the R face, c

_{f}be a 90 degree clockwise rotation of the whole cube centered on the F face, etc.

This convention only defines a notation for 6 of the 24 rotations, but we only need 2 of them as generators for C. For example, we can generate C as C=<c

_{r},c

_{f}> and we can generate M as M=<c

_{r},c

_{f},ν>.

With these generators in mind, we can generate G

_{C}as G

_{C}=<Q,c

_{r},c

_{f}> and we can generate G

_{M}as G

_{M}=<Q,c

_{r},c

_{f},ν>.

G

_{C}can be realized on a physical cube by making standard quarter turns plus rotating the cube as a whole in your hands. G

_{M}cannot be realized on a physical cube (at least, not "really") because it includes improper symmetries, namely the reflections. But you can realize G

_{M}"virtually" on a physical cube by holding your physical cube up to a mirror.

G

_{C}is 24 times larger than G, and G

_{M}is 48 times larger than G.

If you use, for example, an S

_{48}model for the cube, then each of G, G

_{C}, and G

_{M}is a subgroup of S

_{48}. If you use an S

_{24}xS

_{24}model for the cube, then each of G, G

_{C}, and G

_{M}is a subgroup of S

_{24}xS

_{24}. Etc.

Of course, your model for G

_{M}certainly seems correct as well. And if I am understanding your note properly, your model for for G

_{M}provides a convenient framework for proving that any outer automorphism of G is an inner automorphism in some supergroup of G. In other words, for any outer automorphism of G, there exists an element h that is not in G but that is in a supergroup of G such that the automorphism in question may be expressed as G^h.

Jerry

### What I find difficult about t

What I find difficult about this whole subject is that certain things are so closely related that it is so easy to get confused. In particular:

cube positions vs. cube permutations

rotations of 3d space applied to a position, vs. 3d rotations applied to a cube permutation

With these generators in mind, we can generate GC as GC=<Q,cr,cf> and we can generate G_{M}as G_{M}=<Q,cr,cf,v>.

Here you implicitly assume there is a way to multiply an element of G (or Q) and an element of M (or cr,cf,v). Later on you mention considering them as permutations of the 48 facelets, and in that context the multiplication is indeed obvious.

However, it would be better to use S_{54}, i.e. include the centre facelets. Otherwise a 4-spot pattern would be considered the same as a 180 degree cube rotation.

Of course, your model for GM certainly seems correct as well. And if I am understanding your note properly, your model for for GM provides a convenient framework for proving that any outer automorphism of G is an inner automorphism in some supergroup of G.

Exactly. That is really the only point of doing it that way.

In fact, my model is not much different from yours. I constructed an external semi-direct product, whereas you are constructing an internal semi-direct product. In my case G and M are groups related only by how M acts on G and I construct an external supergroup to contain them. In your case you explicitly state how M acts on G by modelling them internally as subgroups if S_{54}.

A nice way to think of this is as follows. G_{M} is really just the set GM = {gm|g in G, m in M}. You need a product rule to make it a group, so you need to know what happens when you multiply the elements g1m1 and g2m2. To get the result g1m1g2m2 to lie in GM, you need to somehow swap m1 and g2. This is where you need to know how M interacts with G.

We have for example that c_{U}q_{F} = q_{R}c_{U}, where the q are quarter turns of faces and c_{U} is a quarter cube rotation around the U axis. Any m1 can be 'pulled through' any g2, and in the process m1 doesn't change though g2 does. In other words, m1g2 = g3m1 for some g3 in G. The product rule then becomes

g1m1.g2m2 = g1g3.m1m2 where g3= m1g2m1'.

(This is slightly different to what I had in the previous post cause I had M and G the other way around).

This is where the conjugation comes in, and g3 is simply g2 acted upon by m1 through conjugation. Also G is normal in GM because any element m can be pulled though, mG=Gm.

Jaap's Puzzle Page:

http://www.geocities.com/jaapsch/puzzles/

### S54 vs. S48

However, it would be better to use SI don't think S_{54}, i.e. include the centre facelets. Otherwise a 4-spot pattern would be considered the same as a 180 degree cube rotation.

_{54}vs. S

_{48}has anything to do with it. Or more correctly, it comes back to your original point about positions vs. permutations. I find the issue confusing on an almost daily basis, and I've been doing this stuff for over 20 years.

With respect to your 4-spot example, the example I like to use is the maneuver RL' on the 2x2x2 cube. Suppose you start with a solved 2x2x2 cube, perform the maneuver RL' on it, toss it to somebody, and ask them to solve it. You would be expecting them to solve it with LR', but instead the person simply tells you that it's already solved. And they would be right. Yet in a computer model or in a mathematical model of the puzzle, it would not be solved. That is, in the computer model the "numbers would no longer be lined up right". It would no longer be the case that 0 was mapped to 0, that 1 was mapped to 1, etc.

The same thing is true of the 4-spot pattern on a 3x3x3 cube. In a computer or mathematical model, you don't need the face centers to tell if the "numbers are lined up right". So for a computer model or mathematical model, an S

_{54}model does not contain any information that is not just as available with an S

_{48}model. But on a physical cube, you do need the face centers. Otherwise, all 24 rotations of a particular position appear to be equivalent.

As a further example, take a 3x3x3 cube and remove the color tabs from all the edges. What you have left is the corners of the 3x3x3. Now, remove the color tabs from the face centers. What you have left is equivalent to the 2x2x2. So the corners of the 3x3x3 is not the same group as is the 2x2x2. It is tricky to model cubes that don't have face centers, as witness the recent discussion about modeling the 2x2x2 with GAP.

Jerry

### The same thing is true of the

The same thing is true of the 4-spot pattern on a 3x3x3 cube. In a computer or mathematical model, you don't need the face centers to tell if the "numbers are lined up right". So for a computer model or mathematical model, an S54 model does not contain any information that is not just as available with an S48 model.

I still disagree, but I probably just misconstrued what you said.
Yes, S_{48} is enough to fully hold a model of the cube.
But no, in this model cube rotations cannot be modelled purely as facelet permutations in the same way as moves. Maybe that wasn't what you were trying to convey, but that is what your post seemed to indicate.

Let me try to explain this.

S_{48} models the permutation of facelets. Each facelet obviously has a home location, and when they are in their home locations, you have the identity perm.

Any move or move sequence moves the facelets from their home positions, and is therefore a permutation in S_{48}.

Any cube rotation however not only moves the facelets, it changes their home locations as well, i.e. is a recolouring. Within the S_{48} model, you can only do this by conjugation m'g m where g is the current permutation, and m is the facelet permutation of the cube rotation/reflection. It is this pre-multiplication that in effect re-defines what the home locations are of the facelets that g permutes.

To take the 4-spot example, it has the same facelet permutation as a 180 degree cube rotation around a face. Applying the 4 spot perm m to a permution g gives gm (I multiply on the right), whereas applying the cube rotation to g would give m'gm instead.

What I said about using S_{54} (or S_{24}xS_{24}xS_{6}) was not correct, or at least incomplete. In this model you can let the cube rotations/reflections be ordinary permutations, and apply them just like moves (i.e. no conjugation needed). The extra S_{6} keeps track of the centre facelet permutation, and thus contains all the info on which cube rotations/reflections have been applied.

Unfortunately, as you point out, this model has 48 solved positions. What we really want to use is S_{54}/M, i.e. factor out those cube rotations/reflections at the end.

In effect, you save all the pre-multiplications of all the cube rotation/reflection conjugations, and do their combined effect at the end.

Jaap's Puzzle Page:

http://www.geocities.com/jaapsch/puzzles/

## innie or outie

>Are the two definitions equivalent?

I think so.

From your description it is not intuitively obvious that such a G

_{M}exists. However, what should be possible is to construct a group with the properties we need.We have the cube group G, and symmetry group M. There is also a group action, where every element of M acts on the set G, in such a way as to preserve its group structure.

Define a group H as an external semi-direct product of G and M as follows:

H = { (m,g) | m in M, g in G }

with the product rule:

(m1,g1)x(m2,g2) = (m1m2, g1^m2' g2 )

where g1^m2' is the result of the action of m2' on g1.

In actual fact, you can consider g3 to be equal to m2'g1m2, though that would really be circular reasoning because we haven't defined any kind of multiplication between elements of M and of G yet.

Check this is associative:

[(m1,g1)x(m2,g2)]x(m3,g3) = (m1m2, g1^m2' g2 )x(m3,g3) = (m1m2m3, (g1^m2' g2)^m3' g3 )

(m1,g1)x[(m2,g2)x(m3,g3)] = (m1,g1)x(m2m3, g2^m3' g3 ) = (m1m2m3, g1^(m3'm2') g2^m3' g3 )

And indeed (g1^m2' g2)^m3' = (g1^m2')^m3' g2^m3' = g1^(m3'm2') g2^m3'

In this group G is isomorphic to {(e,g)|g in G} and M is isomorphic to {(m,e)|m in M}, both subgroups of this external semidirect product H that we constructed from G and M.

If we now identify G and M with those subgroups of H, then we find that H is actually the G

_{M}we were looking for, and things become very much simpler. It is at this point that you can see that the product rule use above isn't some rabbit pulled out of a hat. If you remove all the comma's and brackets, you simply get:m1g1 m2g2 = m1m2 g3g2

where g3 = m2' g1 m2, an obvious truism that allow us to rewrite any element as a product of an element of M and an element of G.

Inner automorphisms of G are still just conjugations with some fixed element of G, and outer automorphisms of G are now congugations with some element of G

_{M}not in G.Of course there may be other outer automorphisms of G. If there are further outer automorphisms, then they generate some group which we can use as M in the above construction, and so we find that any outer automorphism of G is an inner automorphism in some supergroup of G.

Jaap's Puzzle Page:

http://www.geocities.com/jaapsch/puzzles/