# Why is conjugacy as an equivalence relation special in groups?

The conjugacy transformation seems to be especially special in groups.

Defining a conjugacy equivalence relation and splitting groups into their conjugacy classes is vital in classifying and decomposing groups.

My question is, intuitively, why is the conjugacy equivalence relation special (as opposed to using some other notion of 'equivalence') for groups?

My only current intuition is that, for matrix groups, matrices that differ only by a basis transformation do intuitively represent the same symmetry transformation (just in another basis). However this intuition seems to be tied to matrices, so I was wondering if there was a better, more fundamental, intuition?

• One reason is that inner automorphisms are natural automorphisms. Also, the conjugacy classes reflect the amount of commutativity in the group.
– lhf
Commented Dec 15, 2021 at 11:04
• If we view a group as the fundamental group of a path-connected topological space, which we can always do, then conjugation corresponds to a change of base point. (This isn't "intuition", but means that conjugation arises naturally when we view groups in certain ways.) Commented Dec 15, 2021 at 11:05
• @user1729. What do you mean by 'base point'? Also why can we always consider it in a path-connected space, even for non simply connected groups? Commented Dec 15, 2021 at 11:11
• taking a step back, here's a very elementary observation (implicit in the answer and comments): conjugacy is not just an equivalence relation, it's an automorphism. so if two elements of the group are conjugated, then all their group-theoretic properties are the same Commented Dec 15, 2021 at 11:19
• Okay, intuitively I can see the 'specialness' of the automorphism. Why (like in @Shaun's answer), is conjugacy the only way of obtaining inner automorphisms? Commented Dec 15, 2021 at 11:22

Let's take a group of rotations of a cube.

There are 24 elements in this group, let's take a look at these two:

• A: rotate cube 90 degrees clockwise around x axes
• B: rotate cube 90 degrees clockwise around z axes

These two elements are obviously different, but they look somewhat similar. A person looking at our cube from different direction can even be confused by these descriptions. But for example element "C: rotate cube 120 degrees around this diagonal" is very different from A and from B.

Elements A and B are conjugated - that means $$A = x * B * x^{-1}$$, and this conjugation operation is actually "let's take a look at our group from another direction, $$x$$ defines this another direction".

If we have two persons looking at our cube from different directions, than it may happen that the first person takes our description of element A, second person takes our description of element B, but they would correspond to the same actual rotation of the cube.

Conjugated elements are somewhat indistinguishable, that's why there is equivalence relation between them.

UPDATE:

Actually it's an answer to a question from comment about how $$x$$ and "looking from another direction" are related.

Let's get back to the example with rotations of cube. Imagine, that you sit at the table, the cube is in front of you. Your friend sits at the same table, looks at the same cube from the right. So, the face of the cube which is "right" for you is "front" for him.

He describes one of the rotations of cube: "A: rotate cube 90 degrees clockwise around the axis which goes from left face to right face".

You want to find out what is the description of this rotation in your frame of reference. One way for you to do it would be to:

1. move yourself right, to the seat that your friend occupied
2. make the rotation A as described
3. move back to your original seat
4. look at the current orientation of the cube and describe the rotation which moves the cube from original to current state.

But instead of moving yourself around the cube you can move the cube! So, the steps would be:

1. rotate cube around vertical axes 90 degrees counterclockwise
2. make the rotation A as described
3. rotate cube around vertical axes 90 degrees clockwise

So, in your frame of reference the rotation will be a product of these three rotaions:$$A_c = x * A * x^{-1}$$ where $$x$$ is "rotate cube around vertical axes 90 degrees clockwise".

(remember that $$x * A * x^{-1}$$ corresponds to (step 3) * (step 2) * (step 1) )

• Nice and intuitive answer. For completeness sake, which element $x$ would you need to do this? Commented Dec 15, 2021 at 23:11
• @AccidentalTaylorExpansion Thank you, and that you for a good question. Answer would not fit into comment, so I have included it into my original answer. Commented Dec 16, 2021 at 12:47

Well,

$$G/Z(G)\cong {\rm Inn}(G),$$

where

$$Z(G)=\{k\in G\mid kg=gk\forall g\in G\}$$

is the centre of $$G$$ and $${\rm Inn}(G)$$ is the group of inner automorphisms of $$G$$; that is, for all $$g\in G$$, $$c_g\in{\rm Inn}(G)$$ iff $$c_g: G\to G$$ such that $$c_g(h)=ghg^{-1}$$ for all $$h\in G$$. There are no other inner automorphisms.

• I have a physics background, so I would be very grateful if you could elaborate a bit more about the 'centre' and 'inner automorphisms' Commented Dec 15, 2021 at 11:10
• I have included a definition of the centre, @AlexGower. I have already defined inner automorphisms.
– Shaun
Commented Dec 15, 2021 at 11:12
• So, loosely speaking, is this saying that Inn(G) (which contains all 'invertible basis transformations' that you can do with a $g \in G$) is isomorphic to G/Z(G) (which is essentially G if you set all points k to e)? If so, this statement still seems to rely on some notion of the conjugacy since Inn(G) seems to be defined by it, so why is it special here? Commented Dec 15, 2021 at 11:20
• I think my actual question is, why are there no other inner automorphisms? Commented Dec 15, 2021 at 11:22
• Well, then, to see a proof of the isomorphism above, see Theorem 9.4 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)".
– Shaun
Commented Dec 15, 2021 at 11:24

The importance of group conjugation comes down to the following.

Conjugation is an automorphism of a group. In many fields of mathematics, automorphisms are of high importance as they are in essence the symmetries of the underlying structure. Among all automorphisms of a group, the conjugations are special, since their description $$c_g(x) = gxg^{-1}$$ works uniformly in all groups and requires only basic group arithmetic. In this sense:

1. Conjugations are the "obvious" or the "natural symmetries" of a group.

Other anwers hint at the role of conjugation in group actions. In a nutshell, it can be stated as:

1. Conjugation is the formula for the "change of basis" in general group actions.

While this interpretation is certainly important to have in mind, I still think the foremost reason for the significance of conjugation is № 1. Because it works for general groups, without the need of introducing any additional structure (like the group acting on some set).

• Are there other alternative automorphisms we could use though, that still only require basic group arithmetic? i.e. is conjugation the only choice for a nice automorphism using just basic group arithmetic, or is it just a nice choice out of many possibilities? (If the latter, does that just mean it is convention?) Commented Dec 17, 2021 at 11:16
• @AlexGower Often enough, the conjugations are the only automorphisms of a group (Look at en.wikipedia.org/wiki/List_of_finite_simple_groups and read the entries of "outer automorphism group", if you find "Trivial" then there are only inner automorphisms, in other words only conjugations.) So in general, there are no other "obvious" automorphisms. Commented Dec 17, 2021 at 15:31
• I mean, it's still conceivable that there is another class of "uniformly definable" automorphisms, which sometimes coincides with the conjugations and sometimes yields something new. However, I'm not aware of such automorphisms, and I don't think they exist. Maybe it is possible to make this statement more precise by means of category theory, but this is beyond my knowledge. Commented Dec 17, 2021 at 15:36
• Oh I see. This is a very nice explanation. So it sounds like the choice of using conjugation is a bit stronger than them just being the most 'obvious'. It sounds like you really would have to define a quite bizarre object (to coincide with conjugations for groups with no outer automorphisms, but be different for other groups) to use something else Commented Dec 17, 2021 at 15:38
1. In group theory group action is a very important tool .

Group acting on itself is one type of important group action.

Group acting on itself by Left multiplication gives us an way to see the famous and single most important theorem in group theory, Cayley's theorem (Every group $$G$$ can be embedded inside a permutation group $$A(S)$$ for an appropriate $$S$$ )

Another important action group acting on itself by conjugation.

$$\phi:G×G\to G$$ by

$$\phi(g, s) =gsg^{-1}$$

Then, $$orbit(s) =O(s)=\{g\cdot s : g\in G \}$$

Define a relation $$\sim$$ on $$G$$ ,

$$s_1 \sim s_2$$ if $$s_1 =gs_2 g^{-1}$$ for some $$g\in G$$

Then, $$\sim$$ is an equivalence relation, is called conjugacy relation.

And, $$[s]=O(s)=$$= conjugacy class of $$s$$

It's important to prove class equation and Sylow's theorem, important tool to analyze group of finite order.

1. The importance of any equivalence relation is that it partitioned the set into disjoint equivalence classes.

2. $$\phi:G\to G$$ defined by

$$\phi_g(s) =gsg^{-1}$$ is an inner automorphism.

And, if $$H\le G$$ and $$H$$ is stable under conjugation i.e $$\phi(H) =H$$ , then this $$H$$ is an important type of subgroup of $$G$$ , called normal subgroup.