I am learning group theory, and while learning automorphisms, I came across conjugation as an example in many textbooks. Though the definition itself, (and when considering the case of abelian groups), it seems pretty innocent, I have to admit that I have no intuition about whats happening, or why even such a map is important. Did anyone feel this way when you learnt it? Any examples/references that you have found useful when you learnt it?

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    $\begingroup$ An example of the top of my head where conjugation arises is related to base changes in linear algebra. Say that a linear map is represented by matrices $N$ and $M$ in two different bases, and that the base change matrix is $P$, then the relationship between $N$ and $M$ is $N = P^{-1}MP$. $\endgroup$ Nov 26 '10 at 16:55
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    $\begingroup$ Every group can be seen as the fundamental group of some connected topological space. Conjugation corresponds to a change of the base point of this space. $\endgroup$
    – user1729
    Nov 4 '12 at 15:48
  • $\begingroup$ See figure 11 here, related to the pictures in the answer here (because the "Lie algebra version" [i.e. the differential] of $xyx^{-1}y^{-1}$ is the Lie bracket) ! $\endgroup$
    – Watson
    Nov 28 '18 at 21:15

One way to think about conjugation is as a generalization of changing coordinates when rewriting matrices or, from a physical point of view, as the process of "seeing a group from a different perspective."

Let's make this precise. Let $G$ be a group and let $X$ be a set on which $G$ acts faithfully. For example, if $G$ is the Euclidean group of isometries of the plane, $X$ could be... well, the plane. Let's say we have a reasonably concrete description of what a particular element $g \in G$ does to $X$. For example, if $G, X$ are as above, then perhaps $g$ is a rotation counterclockwise by $\theta$ centered at the origin.

Given that description, what does the element $hgh^{-1}$ do? Well, this is actually quite simple: in the description of $g$, replace all the elements $x \in X$ that occur by the corresponding elements $hx \in X$. For example, if $h$ is a translation by a vector $v$, then $hgh^{-1}$ is a rotation counterclockwise by $\theta$ centered at $v$ instead of the origin.

In other words, what conjugation does in terms of group actions (and it is always a good idea to think of groups in terms of their actions) is it corresponds to a "change of coordinates" on the underlying set $X$. This is a basic reason conjugation and group theory are important in understanding Newtonian mechanics (where $G$ is the Galilean group) or relativity (where $G$ is the Lorentz group) or really much of modern physics in general: in physics groups are always studied because of their actions and we always want our concepts and equations to be invariant under these actions. (For example, mass and charge are invariant concepts. The mass or charge of an object doesn't change when you rotate or translate it.)

This gives a very intuitive definition of a normal subgroup: it's a subgroup that "looks the same from every perspective." For example, the subgroup of translations in the Euclidean group is always normal because the description "$g$ is a translation" is the same from every perspective (that is, it's invariant under conjugation). It's a good exercise to look at some groups you're familiar with and see if you can identify which subgroups are normal based on this principle. (This principle underlies the importance of normal subgroups in Galois theory as well.)

  • $\begingroup$ Wouldn't that give an "intuitive definition" of a characteristic subgroup? $\endgroup$ Nov 26 '10 at 17:14
  • $\begingroup$ @Mariano: I have to admit I do not have a good conceptualization of how to think of a group G acting on a set X while G is also being acted on by Aut(G). So I tend to ignore outer automorphisms (except when G is abelian). $\endgroup$ Nov 26 '10 at 17:17
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    $\begingroup$ Heh. To be honest, I think that it simply does not mean anything to look for "intuition behind conjugation". Conjugation is conjugation, and that's that. At most, one can ask for interesting examples where conjugation is meaningful, so as to familiarize one self with the various avatars of the notion, but I tend to think that people asking for intuitions in situations like this are both misguided and in for a disappointment! $\endgroup$ Nov 26 '10 at 17:43
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    $\begingroup$ @Mariano: strongly disagree. I think it is never misguided to ask for intuition. $\endgroup$ Sep 2 '11 at 13:08
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    $\begingroup$ @Ben: Your and Mariano's comments aren't actually contradictory. The problem is the presumption is that there is intuition "behind" conjugation, rather than intuition about its applications, and intuition derived from working with it directly. Really, I think the idea there is intuition "behind" some abstract notion is an illusion derived from the situation where you only understand one application of the notion. $\endgroup$
    – user14972
    Oct 17 '11 at 8:59

This is a belated follow-up to Qiaochu's nice answer and Mariano's comments on characteristic vs normal subgroups. I cannot yet comment on answers, so I thought I'd expand this to an answer.

One of the classic examples of normal but non-characteristic subgroups are the factors of a direct product. The factors are obviously normal, but they are non-characteristic because they are not invariant if you interchange the factors. This interchange defines an outer automorphism, or an external symmetry.

You can make Qiaochu's sketch of an "intuitive definition" closer to a real definition if you draw a distinction between internal and external symmetries. The question is how you can make that distinction intuitive by itself. Here's my attempt:

Imagine a space with symmetry group $\mathbb Z^2$. Let's call it lattice world. A denizen of lattice world has position but no orientation. This last part is crucial. It means that even though we can look at lattice world from our privileged outsider's perspective, tilting our heads 90 degrees and thereby interchange latitudes and longitudes (showing that the $\mathbb Z$ factors are non-characteristic), a denizen of lattice world has no orientation and therefore cannot turn, so that perspective is inaccessible to him: a change of perspective in his world involves only translations.

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    $\begingroup$ +1 Fully agree. An automorphism is "looking at the group from a different persective." An inner automorphism is when this is a perspective the group itself can take. If $O$ is the octahedral group, $Aut(O)$ is entirely inner because every automorphism corresponds to looking at the octahedron from a different angle, which is exactly what the group elements do. The automorphisms all happen in the language of the group itself. For the quaternion group $H$, we with our birds-eye view can permute $i,j,k$ arbitrarily; but the group itself does not have the power. $\endgroup$ Sep 2 '11 at 13:29
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    $\begingroup$ ...and all of a sudden, it makes complete sense why inner and outer automorphisms are called that. Thank you for the great examples. $\endgroup$ May 5 '16 at 11:21

One example of conjugation you probably already know from linear algebra is change of base.

The linear transformation A that takes (x,y) to (3x−y,2x) is reasonably nice and can be represented by the matrix $$A = \begin{pmatrix}3&2\\-1&0\end{pmatrix}$$

In the basis u=2x−y, v=x−y, Au = 2Ax − Ay = 2(3x−y) − 2x = 4x − 2y = 2u and Av = Ax−Ay = (3x−y) - 2x = x−y = v. So in this basis, the linear transformation is incredibly nice and can be represented by the matrix $$\tilde A = \begin{pmatrix}2&0\\0&1\end{pmatrix}$$

If we take the matrix $$B = \begin{pmatrix} 2&1\\-1&-1\end{pmatrix}$$ then $B^{-1} A B = \tilde A$ expresses the matrix in the new basis defined by B, (u,v) = (2x-1y,1x-1y). Note that Bx = u, and By = v.

In some sense this is what conjugation does in general: it changes the basis on which your group acts. If you conjugate by g, it changes the way G acts by converting the old basis to the a new basis already acted on by g.

A good example to work through to see this conjugation in a new context is with permutations. If a permutation A takes 1,2,3 to 3,2,1 and the permutation B takes 1,2,3 to 1,3,2, then the conjugated element $B^{-1} A B$ takes 1,3,2 to 2,3,1. In other words, instead of acting on 1,2,3 it acts on 1,3,2. In cycle notation, you can write this as A=(1,3) and B=(2,3) and then A^B = (1,2). The "3" in (1,3) has been replaced by "2", since B replaces 3s by 2s.

Try to think of groups as acting on something. Either let them be matrices moving vectors around, or permutations moving arbitrary things around. Conjugation just lets you relabel the things you are moving around.

  • $\begingroup$ In general, if $\sigma$ is a permutation and $(x_1, x_2, \ldots, x_n)$ is another permutation, both in $S_m$ ($n\leq m$), then $(x_1, x_2, \ldots, x_n)^{\sigma}=(x_1\sigma, x_2\sigma, \ldots, x_n\sigma)$. This generalises, as an arbitrary permutation $\phi$ is a product of disjoint cycles of the form $(x_1, \ldots, x_{n_i})$ for variable $i$. This is very neat! $\endgroup$
    – user1729
    Oct 17 '11 at 11:11
  • $\begingroup$ (Sorry, I hadn't noticed that this post was the best part of a year old - this thread had just floated to the top of the stack...) $\endgroup$
    – user1729
    Oct 17 '11 at 11:28
  • $\begingroup$ Shouldn't the matrix rows for $A$ be 3, -1, and 2, 0? $\endgroup$
    – user50229
    Apr 24 '13 at 9:58

There are several reasons:

  1. Conjugation is a way of measuring, how commutative or otherwise your group is. Namely $aba^{-1} = b\Leftrightarrow ab=ba$. Since abelian groups are much easier to understand than arbitrary ones, this measure is important.

  2. Conjugation is more or less the only known way of obtaining automorphisms of any given group without knowing anything further about the group (of course, these automorphisms can all turn out to be trivial if your group is abelian).

  3. You have to bear in mind that normal subgroups and the action of conjugation were first introduced by Galois, because for a subgroup to be normal was exactly the criterion that he needed for an automorphism of a field to give an automorphism of a subfield upon restriction. This comment might be slightly over your head if you are only learning groups at the moment, but bear in mind that historically, the notion of normal subgroups arose in the context of interpreting groups as symmetries of field extensions, when Galois developed a theory to decide whether a given polynomial was solvable by using elementary operations of arithmetic and roots.

  • $\begingroup$ That's an amazing answer, thank you so much! $\endgroup$
    – mur_tm
    Jun 6 '20 at 16:54

My intuition goes back to permutations. Namely, conjugating everything by some group element can be identified with some permutation of the group.

As for the importance of conjugation, one easy answer is that it gives insight into the commutative portions of the group. If $x$ is closed under conjugation by $g$, that is $x = gxg^{-1}$, then we have $xg = gx$, so the elements commute. The set of elements that are closed under conjugation by all group elements forms the center subgroup, while any subgroup that is closed under conjugation by all group elements is a normal subgroup. These are fundamental to understanding the structure of a given group.

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    $\begingroup$ Hi, Hans. I was going to say something similar. Especially, one should understand what conjugation means in symmetric groups: in this case, it is literally just expressing things in terms of a relabelling (i.e., bijection) of the elements of the set. $\endgroup$ Apr 16 '11 at 22:46

I was thinking about Qiaochu's answer and Mariano's comment about characteristic subgroups, and I think I have something of an idea of why inner automorphisms should be distinguished.

Let $G$ be a group. We can consider it as a category with one object that we arbitrarily call $*$, in which the morphisms are the group elements and $1_*$ is $G$'s identity. Let $F:G\to\rm Set$ be a functor and denote $X=F*$ the target object. Then group elements are taken to maps in $\hom(X,X)$; since the elements in the group are invertibe these morphisms will be isomorphisms so this is essentially a homomorphism $G\to\mathrm{Aut}(X)$. Effectively $F$ encodes a (left) group action on $X$. (A right action would be encoded with a contravariant functor.)

Since groups abstract the idea of symmetry through their potential actions, it's arguably natural to think of the collection of different ways (in particular the transformations between them) of using the abstract group to concretely realize symmetry (group actions) as "symmetry of symmetries."

If $F,L:G\to X$ are functors represented by the homomorphisms $\alpha,\beta:G\to\mathrm{Aut}(X)$ resp., then a natural transformation between them is an isomorphism $\eta:X\to X$ such that $\alpha\circ\eta=\eta\circ\beta$. That is, two group actions are naturally isomorphic if they are conjugate via $\mathrm{Aut}(X)$.

In particular, a given action $\alpha:G\to\mathrm{Aut}(X)$ provides us with many set automorphisms of $X$ and so conjugating by these, we have natural transformations $(g\mapsto\alpha(g))\mapsto(g\mapsto\alpha(hgh^{-1}))$ for every element $h\in G$. Ultimately then, $G\to\mathrm{Inn}(G)\hookrightarrow\mathrm{Aut}(F)$, where the aut group here tacitly comes from the functor category $\mathcal{C}^G$, allows one to use $G$ to encode its own symmetry via conjugation.


Let $ G $ be a group. With no further specifications, can we form any interesting isomorphisms from $ G $ to $ G $ ?

For each $ g \in G $, we can try to form an isomorphism $ \varphi_g $ from $ G $ to $ G $ using $ g $.

Let's try maps of the form $ t \mapsto g^k t^l g^m $ $(k,l,m \in \mathbb{Z})$. We want $ g^k (t_1 t_2)^l g^m $ $ = g^k t_1^{l} g^m g^k t_2^{l} g^m $ to hold for all $ t_1, t_2 \in G $, and this can be made to happen by setting $ \{ k = (-m), l = 1 \} $. Maps $ t \mapsto g^{-m}tg^{m} $ are bijections too, giving us a family $ \{ t \mapsto g^{-m}tg^{m} : g \in G, m \in \mathbb{Z} \} $ $ ( = \{ t \mapsto g^{-1}tg : g \in G \} ) $ of isomorphisms from $ G $ to $ G $. These isomorphisms are called conjugations.

[ We could've tried, instead of $ g^k t^l g^m $, some longer more general string of $ g $s and $ t $s, but I can't seem to get any more isomorphisms this way ]


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