This definition is fairly mysterious at first. It becomes a bit less mysterious when you learn that N is normal in G precisely when you can "mod out" G by N and get a new group G/N. Guys in G/N are equivalence classes of guys in G. Two guys in G, say g and g', define the same guy in G/N precisely when g' = gn for some n in N. The condition that N be normal is precisely what we need to get a well-defined way of multiplying these equivalence classes!

But what are normal subgroups *like?* How can you just look
at a subgroup and guess if it's normal?

Most people don't have the fortitude to think about this question very hard! Shawn Fitzpatrick did, and he sent me an email about it:

From: Shawn Fitzpatrick

To: John Baez

Subject: What's a normal subgroup?

Date: Tue, 08 Feb 2005 13:45:57 -0500

Hi there. My name's Shawn Fitzpatrick. I'm currently in the process of trying to come to grips with a few modern algebra concepts, and I came across your website, http://math.ucr.edu/home/baez/week155.html.

From the website, I got the impression that you're a kindred spirit, in so far as you seem to crave a visual understanding of abstract concepts. Granted, you seem to be leagues ahead of me, but I've made an educated guess that you enjoy providing guidance for those wrestling with mathematical concepts. Hopefully you won't mind taking time out of your busy day to answer a couple of questions I had.

Outline:

1. Nature of my questions

2. Examples of normal (and abnormal) subgroups

3. My attempted intuitive definition of "normal"

4. Counterexample to the definition?

5. My questions

## 1. Nature of my questions

I'm trying to come to terms with the concept of a "normal subgroup." I've read the definition a million times, but still feel like I lack an intuitive understanding of what normal subgroups actually are. For that reason, I recently adopted a strategy of trying to track down as many examples as I could (preferably visual) and then hoping that an intuitive pattern would emerge.## 2. Examples of normal (and abnormal) subgroups

Here's the progress I've made so far on my "normal subgroup safari"...Groups:

Normal subgroups of the above groups:

- group of rotational symmetries of a tetrahedron
- group of rotational symmetries of a cube (which is effectively the same as that of an octahedron)
- group of rotational symmetries of an icosahedron (which is effectively the same as that of a dodecahedron)

- The group of all rotational symmetries of the tetrahedron such that each edge get mapped either onto itself or onto the opposing edge. (This group of 4 rotations is isomorphic to Z/2 x Z/2 and is a normal subgroup of group 1 above.)

- a. The group of all rotational symmetries of the cube such that each edge gets mapped to a parallel edge. (This group of 4 rotations is also isomorphic to Z/2 x Z/2 and is a normal subgroup of group 2 above.)
b. The group of all rotational symmetries of the cube such that the axis of rotation either passes through the center of 2 opposing faces or through 2 opposing vertices. (This group of 12 rotations is isomorphic to A

_{4}and is a normal subgroup of group 2 above.)- Group 3 is simple, and hence has no normal subgroups.
Subgroups that aren't normal (aka "abnormal" subgroups)

- Pick any of the 4 axes of the tetrahedron which pass through a vertex and the center of a face. Then consider the group = {e, 120 clockwise rotation, 120 counter-clockwise rotation}; This subgroup of Group 1 above, unfortunately, has 3 other conjugates, 1 corresponding to each axis. Since all 4 subgroups are conjugate to each other, relative to Group 1 above, none of them is normal.

- Pick any of the 3 axes of the cube which pass through 2 opposing faces. Consider the group = {e, 90 rotation, 180 rotation, 270 rotation}. Although this is a subgroup of group 2 above, it has 2 conjugates, 1 corresponding to each axis. Hence, sad as it is, none of these 3 subgroups are normal.

## 3. My attempted intuitive definition of "normal"

OK, so, obviously I've limited my examples to groups of rotational symmetries on polytopes (more specifically, polyhedrons). However, as far as these examples are concerned, a pattern seems to have emerged. Namely, a normal subgroup of a given group of rotations is effectively a subgroup which doesn't "look like" any other subgroup. For example, the group of 4 rotations of a cube along the x-axis basically "looks the same" as the group of 4 rotations of a cube along the y-axis, or along the z-axis for that matter. Hence, none of them is unique, so none of them are "normal." On the other hand, consider the group of all 4 of the 180 degree rotations of the cube around one of the 3 axes passing through the centers of opposing faces. Since there is no other subgroup of rotational symmetries of a cube that contains 4 180 degree rotations, this subgroup is unique and hene "normal."In other words...

Let N be a subgroup of a given group G of rotational symmetries on an object. N is a normal subgroup iff:

For every element n of N, there is no element in G which both lies outside of N and which "looks exactly like n does" if you view the polytope from the appropriate angle.This intuitive definition is based on the fact that, where symmetry is concerned, conjugation of a rotation corresponds to changing your vantage point.## 4. Counterexample to the definition?

Counterexample to the intuitive definition (???):The quaternion group of order 8 has 3 subgroups which basically "look like" each other, namely {1,i,-1,-i}, {1,j,-1,-j}, and {1,k,-1,-k}. Despite the fact that these 3 groups are structurally identical, each of them is a normal subgroup of the quaternion group.

Granted, this isn't necessarily a counterexample to the above definition, because the quaternion group doesn't necessarily match the criteria in my intuitive definition of normalness, namely that of "symmetry." The problem is that I don't know. If I assume that there exists no polytope such that the group of rotational symmetries is either isomorphic to or contains the quaternion group, then my intuitive definition is safe. But what if the quaternion can be represented as a group of rotational symmetries in 4 dimensions?

So this brings to my questions (finally).

## 5. My questions:

- Do you agree that, assuming that G is a group of rotational symmetries, a "normal subgroup of G" is basically the same thing as a subgroup of G that doesn't "look like" any other subgroup?
- Can the quaternion group or order 8 be represented as a group of rotational symmetries in 4 dimensions? (On your webpage, there seemed to be an inferred connection between the symmetries of the 4th dimensional 24-cell and the quaternion group)
- Can you suggest an alternative intuitive definition of a "normal subgroup?"
Thank you immensely for taking the time to read through my email; I would appreciate any guidance you could give me to getting a better handle on group theory.

Here's my reply:

From: John Baez

To: Shawn Fitzpatrick

Subject: Re: What's a normal subgroup?

Date: Tue, 8 Feb 2005 11:50:39 -0800 (PST)

Hi -

I'll answer your question but cc all the material you wrote to my friend James Dolan, since he and I have talked a lot about the conceptual significance of normal subgroups, using some examples very much like the ones you give. In fact, since I'm going to write a

lotof stuff, I'll also cc it to a bunch of other people who may be interested.These are good examples for anyone studying the concept normal subgroup.2. Examples of normal (and abnormal) subgroups Here's the progress I've made so far on my "normal subgroup safari"... Groups: 1) group of rotational symmetries of a tetrahedron 2) group of rotational symmetries of a cube (which is effectively the same as that of an octahedron) 3) group of rotational symmetries of an icosahedron (which is effectively the same as that of a dodecahedron)You'll note that none of these subgroups is defined by picking aNormal subgroups of the above groups: 1) The group of all rotational symmetries of the tetrahedron such that each edge get mapped either onto itself or onto the opposing edge (This group of 4 rotations is isomorphic to Z/2 x Z/2 and is a normal subgroup of group 1 above. 2a) The group of all rotational symmetries of the cube such that each edge gets mapped to a parallel edge. (This group of 4 rotations is also isomorphic to Z/2 x Z/2 and is a normal subgroup of group 2 above.) 2b) The group of all rotational symmetries of the cube such that the axis of rotation either passes through the center of 2 opposing faces or through 2 opposing vertices. (This group of 12 rotations is isomorphic to A_{4}and is a normal subgroup of group 2 above.) 3) Group 3 is simple, and hence has no normal subgroups.specificvertex, edge, or face of the solid, and then asserting something about it. You can define them all by saying something is true foreveryfeature of a given type.You did this quite clearly in 1) and 2a), though you used the word "each" - no big deal. In 2b) you could do this using the fact that there are two ways to choose vertices of a cube that are the vertices of a regular tetrahedron fit snugly inside it. You could define the subgroup in 2b) by saying "both tetrahedra in the cube are mapped to themselves". (This is why this subgroup is isomorphic to the group of symmetries of a tetrahedron.)

Note that all these groups are defined by first picking aSubgroups that aren't normal (aka "abnormal" subgroups): 1) Pick any of the 4 axes of the tetrahedron which pass through a vertex and the center of a face. Then consider the group = {e, 120 clockwise rotation, 120 counter-clockwise rotation}; This subgroup of Group 1 above, unfortunately, has 3 other conjugates, 1 corresponding to each axis. Since all 4 subgroups are conjugate to each other, relative to Group 1 above, none of them is normal. 2) Pick any of the 3 axes of the cube which pass through 2 opposing faces. Consider the group = {e, 90 rotation, 180 rotation, 270 rotation}. Although this is a subgroup of group 2 above, it has 2 conjugates, 1 corresponding to each axis. Hence, sad as it is, none of these 3 subgroups are normal.specificfeature of the solid (an axis, for example) then asserting something is true about it. Picking a different feature of the same type (e.g. a different axis) can yield a different subgroup.All this is very good; I've tried to make it a bit more precise in a way that generalizes nicely.3. My attempted intuitive definition of "normal" OK, so, obviously I've limited my examples to groups of rotational symmetries on polytopes (more specifically, polyhedrons). However, as far as these examples are concerned, a pattern seems to have emerged. Namely, a normal subgroup of a given group of rotations is effectively a subgroup which doesn't "look like" any other subgroup. For example, the group of 4 rotations of a cube along the x-axis basically "looks the same" as the group of 4 rotations of a cube along the y-axis, or along the z-axis for that matter. Hence, none of them is unique, so none of them are "normal." On the other hand, consider the group of all 4 of the 180 degree rotations of the cube around one of the 3 axes passing through the centers of opposing faces. Since there is no other subgroup of rotational symmetries of a cube that contains 4 180 degree rotations, this subgroup is unique and hene "normal."Right! Very good!!!In other words... Let N be a subgroup of a given group G of rotational symmetries on an object. N is a normal subgroup iff For every element n of N, there is no element in G which both lies outside of N and which "looks exactly like n does" if you view the polytope from the appropriate angle. This intuitive definition is based on the fact that, where symmetry is concerned, conjugation of a rotation corresponds to changing your vantage point.These subgroups are isomorphic as groups. However, there's another sense in which they don't "look alike". If you take two of them, say H and H', there's no element of the quaternion group such that4. Counterexample to the definition? Counterexample to the intuitive definition (???): The quaternion group of order 8 has 3 subgroups which basically "look like" each other, namely {1,i,-1,-i}, {1,j,-1,-j}, and {1,k,-1,-k}. Despite the fact that these 3 groups are structurally identical, each of them is a normal subgroup of the quaternion group.gHg

^{-1}= H'Of course this is because they're normal. But, conceptually, if gHg

^{-1}= H' were true, this would be saying that H' was just like H "viewed from another vantage point" - where now "changing vantage point" is done not by conjugating with a rotation, as it was in your previous examples, but by conjugating with an element of the quaternion group!So, if you think of conjugation as "change of vantage point", your intuitive definition becomes precise and correct.

I think it's also possible to take my attitude: if you think of the quaternion group as the symmetries of some shape (I think a 4d one will work best), the normal subgroups you mention can be defined by in a way that doesn't involve picking a

specificfeature of some type.Of course, this is another way of saying these subgroups don't depend on a choice of vantage point!

It can. Any finite group is a subgroup of the rotation group inGranted, this isn't necessarily a counterexample to the above definition, because the quaternion group doesn't necessarily match the criteria in my intuitive definition of normalness, namely that of "symmetry." The problem is that I don't know. If I assume that there exists no polytope such that the group of rotational symmetries is either isomorphic to or contains the quaternion group, then my intuitive definition is safe. But what if the quaternion can be represented as a group of rotational symmetries in 4 dimensions?somedimension. In the case of the quaternions there should be a very nice 4-dimensional thing whose symmetry group is the quaternion group, and I should know what it is, but I'm blanking out.Well, I hope I've answered them by now.So this brings to my questions (finally).Yes, using the precise notion of "look like" which I described.5. My questions: 1) Do you agree that, assuming that G is a group of rotational symmetries, a "normal subgroup of G" is basically the same thing as a subgroup of G that doesn't "look like" any other subgroup.All 4d regular polytopes are related to the group of unit quaternions, which contains your "quaternion group" as an 8-element subgroup. But the relation is a bit baroque: a bunch of these polytopes not only2) Can the quaternion group or order 8 be represented as a group of rotational symmetries in 4 dimensions? (On your webpage, there seemed to be an inferred connection between the symmetries of the 4th dimensional 24-cell and the quaternion group)havesymmetry groups, their verticesaresubgroups of unit quaternions.My webpage:

http://math.ucr.edu/home/baez/platonic.html

explains how the quaternion group forms the vertices of a cross-polytope, the 4d analogue of an octahedron.

But, the group of symmetries of the cross-polytope is bigger than the quaternion group.

So, I think we'd need to "color" the cross-polytope in some way to get something whose rotational symmetries were precisely the quaternion group.

Done!3) Can you suggest an alternative intuitive definition of a "normal subgroup?"Best,

jb

Here's another interesting email, which clarifies the idea of normal subgroups as subgroups that are preserved under "change of vantage point":

From: Per Vognsen

To: John Baez

Subject: Normal subgroups

Date: Sat, 26 Mar 2005 17:36:52 -0500

Dr Baez,

I've been following your website for years and am a great admirer of your expository work. I just came across your new article on the intuition behind normal subgroups and thought I might have something to add that could be helpful for your readers.

First I'd like to say I really like your "change of vantage point" approach to understanding normal subgroups. My personal piece of intuition has always been that conjugation corresponds to a generalized "change of coordinates" which is very closely related to what you describe. When I was trying to master the concepts of elementary group theory back in high school, the main example of conjugation I had to draw on was the notion of similarity transformation in linear algebra. If A is an invertible linear transformation representing a change of coordinates and T is an arbitrary linear transformation then we may think of ATA

^{-1}as T expressed in the coordinate system defined by A. This is very intuitive: if x is a vector then (ATA^{-1}) x = A(T(A^{-1}x)) which one may read as:"First transform x into the coordinate system defined by A; then transform by T; and finally transform the result back into the original coordinate system".This intuitive description of conjugation transfers to the more general setting of groups as long as you are willing to accept a more general notion of "change of coordinates": multiplication by any element of a group acts as a permutation of the group which we may think of as a "re-labeling" of the group elements, aka a "change of coordinates".Cheers,

Per Vognsen

© 2005 John Baez

baez@math.removethis.ucr.andthis.edu