Intuition behind conjugation in group theory

Solution 1:

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.)

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.