37
$\begingroup$

In classifications of the subgroups of a given group, results are often stated up to conjugacy. I would like to know why this is.

More generally, I don't understand why "conjugacy" is an equivalence relation we care about, beyond the fact that it is stronger than "abstractly isomorphic."

My vague understanding is that while "abstractly isomorphic" is the correct "intrinsic" notion of isomorphism, so "conjugate" is the correct "extrinsic" notion. But why have we designated this notion of equivalence, and not some other one?

To receive a satisfactory answer, let me be slightly more precise:

Question: Given two subgroups $H_1, H_2$ of a given group $G$, what properties are preserved under conjugacy that may break under general abstract isomorphism?

For example, is it true that $G/H_1 \cong G/H_2$ iff $H_1$ is conjugate to $H_2$? Or, is it true that two subgroups $H_1, H_2 \leq \text{GL}(V)$ are conjugate iff their representations are isomorphic? I'm sure these are easy questions to answer -- admittedly, I haven't thought fully about either -- but I raise them by way of example. What are other such equivalent characterizations?

$\endgroup$
  • 1
    $\begingroup$ Are we assuming $H_1$ and $H_2$ normal? $\endgroup$ – goblin Dec 18 '15 at 8:55
  • 1
    $\begingroup$ @goblin: not necessarily. To endow $G/H_1$ and $G/H_2$ with group structures, then yes, I should have added normality in that line; I was being careless. $\endgroup$ – Jesse Madnick Dec 18 '15 at 8:58
  • 5
    $\begingroup$ Note that $H_1$ is normal if and only if there are no other conjugate subgroups to $H_1$. So it's not very interesting to assert that $G/H_1 \simeq G/H_2$ when $H_1$ and $H_2$ are conjugate subgroups, and both normal. $\endgroup$ – Dustan Levenstein Dec 18 '15 at 18:44
  • $\begingroup$ So many good answers in such a short time. Peeps are really up on their algebra! $\endgroup$ – ThomasMcLeod Dec 19 '15 at 4:46
23
$\begingroup$

This answer isn't of the form you asked for, but I'm posting it because it's been very helpful to me. I apologize if this is the sort of answer you're trying to avoid.

Let's start with a very specific situation. Let $G$ be the isometry group of the plane $E$. If you pick a point $x$, the rotations around $x$ form a subgroup $G_x$ of $G$. If you pick two different points $x$ and $y$, you get two subgroups $G_x$ and $G_y$. These subgroups are different, but they don't feel really different, because you can turn one into the other just by changing your point of view: if you shift the plane by an isometry that puts $x$ where $y$ used to be, then $G_x$ becomes the subgroup $G_y$ used to be. In other words, $G_x$ and $G_y$ aren't really different because there's an isometry $g \in G$ that makes the diagram $$\require{AMScd} \begin{CD} E @>G_x>> E \\ @VgVV @VVgV \\ E @>>G_y> E \end{CD}$$ commute. (In this diagram, each arrow stands for a whole set of isometries, with $E \overset{g}{\longrightarrow} E$ denoting the singleton $\{g\}$, and composing arrows means composing all pairs of isometries.)


Now, to be completely general, think of any group $G$ as a group of "allowed symmetries" of some object—in other words, a subgroup of $\operatorname{Aut} X$ for some object $X$. (This is completely general because we can take $X$ to be $G$ itself with the left multiplication action. Perhaps more satisfyingly, we can make $X$ a graph if $G$ is finite, a tree if $G$ is free, an effective Klein geometry if $G$ is a suitable Lie group...)

Once again, even if two subgroups of $G$ are different, they don't feel really different if you can turn one into the other just by looking at $X$ from a different point of view. In other words, two subgroups $H, \tilde{H} < G$ aren't really different if there's an allowed symmetry $g \in G$ of $X$ that makes the diagram $$\require{AMScd} \begin{CD} X @>H>> X \\ @VgVV @VVgV \\ X @>>\tilde{H}> X \end{CD}$$ commute. This, of course, is the definition of conjugacy.


This point of view, by the way, is my favorite motivation for the idea of normality: a normal subgroup is a class of symmetries that doesn't depend on your point of view. If you shift the plane by an isometry, for example, your notion of what counts as a horizontal translation might change, but your notion of what counts as a translation won't. Horizontal translations form a non-normal subgroup, while translations form a normal subgroup.

$\endgroup$
  • $\begingroup$ I like this point of view too, and have been slowly thinking about a similar answer to a different question! But I'm curious: What exactly is going on with this map $X \overset{H}{\to} X$ in your second diagram? $\endgroup$ – pjs36 Dec 18 '15 at 19:24
  • $\begingroup$ Right, @pjs36, I should have said what I meant. I've edited the answer to clarify. Each arrow in the diagram stands for a whole set of maps, with $X \overset{g}{\longrightarrow} X$ denoting the singleton $\{g\}$, and composition means composition of all possible pairs. This mirrors the common notations $gH$ for cosets and $KH$ for products of subgroups. $\endgroup$ – Vectornaut Dec 18 '15 at 19:54
  • $\begingroup$ Thanks! I realized after a bit that that must be what it meant, I just couldn't put it all together at first. I do like the interpretation and connection with $gH = Hg$, very nice. $\endgroup$ – pjs36 Dec 18 '15 at 20:00
  • $\begingroup$ @pjs36, glad you like it! $\endgroup$ – Vectornaut Dec 18 '15 at 20:02
20
$\begingroup$

$H_1$ and $H_2$ are conjugate as subgroups iff $G/H_1$ and $G/H_2$ are isomorphic as $G$-sets.

Edit: Two related settings where this condition shows up are Galois theory and covering space theory. One way to state the classification of (not necessarily connected) covering spaces of a (nice, path-connected) space $X$ is that the category of such covers is equivalent to the category of $\pi_1(X)$-sets. Among these, the transitive $\pi_1(X)$-sets correspond to the connected covers, so we get that connected covers correspond to conjugacy classes of subgroups of $\pi_1(X)$. To get subgroups on the nose you need to pick basepoints in the covers lifting a basepoint in $X$.

Edit #2: Here is another setting where this condition appears, to whet your appetite. If $H_1$ and $H_2$ are isomorphic, then their categories $\text{Rep}(H_1)$ and $\text{Rep}(H_2)$ of linear representations are equivalent. But if $H_1$ and $H_2$ are conjugate, it's furthermore true that we can choose an equivalence between these categories to have the property that the corresponding induction functors $\text{Ind}_{H_i}^G : \text{Rep}(H_i) \to \text{Rep}(G)$ are naturally isomorphic.

$\endgroup$
  • $\begingroup$ Good answer. See here for the relevant definitions. $\endgroup$ – 6005 Dec 18 '15 at 10:27
  • 2
    $\begingroup$ This is probably too similar to warrant a different answer, so I'll post it as a comment on this. If you take a representation of $G$ and restrict it to two abstractly isomorphic subgroups $H_1$, $H_2$ the restrictions may not be isomorphic as representations under any identification, but for two conjugate representations they are. $\endgroup$ – Nate Dec 18 '15 at 18:00
  • 1
    $\begingroup$ I second the comment by Nate. See the importance of "trace of Frobenius" in understanding the statements of various theorems and conjectures arising from the area surrounding the study of Galois representations. $\endgroup$ – tkr Dec 18 '15 at 21:27
  • 1
    $\begingroup$ The statement was asked as a question at math.stackexchange.com/questions/579642 $\endgroup$ – Colin McQuillan Dec 19 '15 at 8:33
17
$\begingroup$

I will not attempt a complete answer, but just as an example, if you are studying finite simple groups $G$ (which lots of people do), then one of the first questions you might ask about $G$ is: what are its maximal subgroups? A complete list would be unmanageably long, whereas a list of conjugacy class representatives provides all of information you would need. In this case, the number of conjugates of a maximal subgroup $M$ is equal to $|G:M|$ so you know exactly how many of them there are.

A list of isomorphism class representatives would not be useless, but would leave you with unanswered questions, such as how many maximal subgroups there are altogether. For example, for ${\rm PSL}(2,7)$, there are just two such $S_4$ and a nonabelian group of order $21$. It turns out that there is a unique conjugacy class of the second of these (it is actually the normalizer of a Sylow $7$-subgroup, so that is clear), but two conjugacy classes of $S_4$, which are conjugate in the automorphism group of ${\rm PSL}(2,7)$, which is ${\rm PGL}(2,7)$.

In fact in general, whenever you have isomorphic but not conjugate subgroups, then you will be interested in knowing whether they are conjugate in ${\rm Aut}(G)$.

$\endgroup$
  • $\begingroup$ You talk about "$G$" in the second paragraph, but do not define it. $\endgroup$ – Pedro Tamaroff Dec 18 '15 at 18:14
  • $\begingroup$ I meant ${\rm PSL}(2,7)$ - I have edited it. $\endgroup$ – Derek Holt Dec 18 '15 at 18:24
13
$\begingroup$

I don't understand why "conjugacy" is an equivalence relation we care about, beyond the fact that it is stronger than "abstractly isomorphic."

An observational fact about conjugacy: things are conjugate if they do "the same thing" relative to different perspectives. For instance, consider:

  • Conjugating a permutation relabels the entries in its two-line notation (top and bottom row), or equivalently relabels the entries in its cycle notation. The ralabelling is according to the permutation one conjugates by.
  • Conjugating a matrix represents change in coordinates. Given a linear transformation, two (ordered) bases, and two corresponding representations of the linear map with respect to the bases as matrices, the two matrices are conjugate by the change-of-basis matrix.
  • Conjugating a loop in a pointed fundamental group by a path yields a loop around a possibly different basepoint obtained by dragging the original loop along said path.
  • Two linear representations are equivalent precisely when they are conjugate by some intertwining aka equivariant isomorphism.

Caveats: the first two examples are about conjugating individual elements of groups, the last two examples may just as easily involve conjugating by things outside of a group in order to move between distinct groups, and the last example involves conjugating a group homomorphism pointwise rather than a subgroup. But the principle is still the same.

The highest calling a group can have in life is to act on something, so it is more than just an abstract algebraic structure isolated from the world. To this end, groups are to group actions as potential energy is to kinetic energy, except with symmetry instead of energy. In this spirit, there should be some action-related interpretation of subgroups which helps us understand why conjugation should be important and why it appears in nature as in the above examples. The idea is that any subgroup is potentially a point-stabilizer.

Point-stabilizers are significant because of the orbit-stabilizer theorem. While usually cast as a quantitative identity, we can instead think of OS as something richer: given any point in a space/set, the act of applying group elements to that chosen point induces a "covering" of the point's orbit by the group itself. The fibers of this map are precisely the cosets of the point's stabilizer. Indeed, under nice conditions, we can say Stabilizer → Group → Orbit is a fiber bundle, so the group is a bunch of copies of the stabilizer parametrized by the orbit.

Where then does conjugation enter into the picture? Well, two point-stabilizers are conjugate by a group element whenever that group element takes the one point to the other. Examples:

  • In the (orientation-preserving component of) Euclidean group of rigid motions of space, the stabilizer of a chosen point is the subgroup of rotations around that point. Two such rotation subgroups are conjugate by the translation which shifts one point to the other.
  • In the affine group of space, the stabilizer of a point are the "linear maps" of the vector space obtained by treating that point as the origin. Two such stabilizers are again conjugate by a translation between points.
  • With wallpaper groups, or higher-dimensional space groups more generally, the stabilizers are called crystallographic point groups, and they are once again conjugate by translations (those in the lattice of such that are symmetries of the original tiling or crystal).
  • In the isometry group of the hyperbolic plane, stabilizers are once again "rotation" subgroups, and they are also again conjugate by elements of a special subgroup acting regularly on the hyperbolic plane by "translations" similar to the above examples, but in this case there is no linear structure at all in the situation. This generalizes to higher dimensions too.
  • In the dihedral group acting on a regular polygon, the point stabilizers are flips across bisecting lines through vertices, two of which are conjugate by the rotation that moves the first vertex to the second. This is a discrete case, of which there is a continuous analogue: the orthogonal group acting on the circle.
  • With an orthogonal group acting on a higher-dimensional sphere, the stabilizer of a vector is canonically isomorphic with the orthogonal group of the vector's orthogonal complement, and these stabilizers are also conjugate by rotations.

Often other subgroups too may be compared or contrasted based on what they do in the group action, and this is reflected by conjugacy. We can see this with some one-parameter subgroups:

  • Given any collection of orthogonal, oriented planes in space and a "speed" associated with each plane, there is a one-parameter subgroup which restricts to rotations of these individual planes at their associated speeds along their orientation. In fact, these are all one-parameter subgroups. Conjugating it by a chosen rotation yields a new one-parameter subgroup: one just applies the chosen rotation to all of the individual orthogonal, oriented planes to get the new collection of planes (whilst keeping the speeds).
  • Beyond the point-stabilizers which act like rotation groups in the hyperbolic plane, there are two other types of actions: those that rotate around a "point at infinity" (so on the boundary in Poincare's disk model) and those that make points flow on either side of a given geodesic. Both types yield one-parameter subgroups. Two "rotation groups" around two different "points at infinity" are conjugate by an element of the aforementioned regular subgroup that moves the one point to the other. Similarly, two "geodesic river flows" are conjugate by the element of the regular subgroup that moves one geodesic to the other.
$\endgroup$
  • 1
    $\begingroup$ The Stabilizer → Group → Orbit fiber bundle picture is lovely. I'm sure someone's tried to show it to me before, but it never clicked until I read this aswer. Thanks for mentioning it! $\endgroup$ – Vectornaut Dec 24 '15 at 22:09
  • $\begingroup$ "The highest calling a group can have in life is to act on something": what a nice phrase this. May I ask if it's due to yourself or else could you point to a reference? Also could you expand on the energy analogy? $\endgroup$ – Alp Uzman Sep 5 '18 at 16:50
  • $\begingroup$ @AlpUzman Due to myself. Generally when groups act on things, we think of the action as some kind of "motion," hence kinetic. An abstract group has the potential to act on things if we so choose to let it. $\endgroup$ – arctic tern Dec 5 '18 at 17:48
  • $\begingroup$ Thanks! I think the energy analogy in this aspect is a little misleading; groups and group actions are as far as i can tell more intimate than different flavors of energy; they are more like "time" and "motion", both of which are intimately related to "change". For instance a differential equation determines a flow ("motion"), whose associated notion of "time" is the additive group of real numbers. $\endgroup$ – Alp Uzman Dec 6 '18 at 20:42
  • $\begingroup$ @AlpUzman There's a difference between potential and actual. When you're talking about a particular group action, it is "actual," but if you're talking about an abstract group (e.g. in any context where one says "let $G$ be a group..." with no assumption of what the group elements actually are or do), it merely has the potential to act in various ways. I agree with your descriptor of "motion," which is why I used the word "kinetic" in the potential vs. kinetic distinction. $\endgroup$ – arctic tern Dec 8 '18 at 22:10
6
$\begingroup$

In classifications of the subgroups of a given group, results are often stated up to conjugacy. I would like to know why this is.

Conjugacy is important in Galois-like bijections: pointed covering spaces of a (nice) pointed space $(X,x)$ are classified by subgroups of the fundamental group of $X$, and conjugacy classes represent different choices of points lying over $x$. Similarly, conjugate subgroups of the Galois group $G$ of an extension $E/K$ correspond to conjugate subextensions, that is, subextensions $F,F'$ such that $\sigma F=F'$ for some automorphism $\sigma/K$. In the lines of what you mentioned, conjugate matrices are those that give isomorphic $F[X]$-module structures on a finite dimensional $F$-vector space $V$. This is essentially what many people learn in a not-so-introductory linear algebra course: the Jordan canonical form is the answer (sometimes) to "when are two matrices (representations) conjugate?", say!

The problem of conjugacy is thus well motivated! A slightly duller example is that conjugacy classes of a group give the dimension of $Z(kG)$ over $k$ and in fact give a canonical representative. I recall having attended a talk where it was vaguely mentioned that conjugacy classes plus something else classify Yetter-Drinfeld modules over the Hopf algebra $kG$. (Add: checking Wikipedia, you want $k=\Bbb C$, and modules are classified by a conjugacy class say $\overline g$ and an irreducible representation of the centralizer of $g$ in $G$).

$\endgroup$
2
$\begingroup$

Conjugacy of subgroups has important meaning in applications to topology and geometry. The vague idea is that homotopy relative to a base point preserves elements and subgroups of fundamental groups, but allowing the base point to move freely only preserves conjugacy classes. Here are some explicit examples.

-- Suppose that $G$ is a torsion free group and that $G = \pi_1(X)$ for some sufficiently nice path connected space $X$ (such as a manifold or CW complex). In this situation the conjugacy classes of infinite cyclic subgroups of $G$ correspond one-to-one with the nontrivial homotopy classes of continuous maps $f : S^1 \to X$ up to precomposition by a self-homeomorphism of $S^1$.

-- For a more geometric version of the above example, suppose furthermore that $X$ is a compact, connected, negatively curved Riemannian manifold. Then the conjugacy classes of infinite cyclic subgroups of $\pi_1(X)$ correspond one-to-one with maps $S^1 \to X$ that are reparamaterized local geodesics (up to precomposition by an isometry of $S^1$).

-- In a more general vein, if $X,Y$ are path connected CW complexes and if both are Eilenberg-Maclan spaces, then the conjugacy classes of subgroups of $\pi_1(Y)$ isomorphic to $\pi_1(X)$ correspond one-to-one with the homotopy classes of continuous maps $f:X \to Y$ modulo precomposition by self-homotopy equivalences of $X$.

$\endgroup$
2
$\begingroup$

It has already been mentioned by arctic tern mainly by giving examples and is implicit in the answers of others, but just to make it more explicit. There is a close connection between the action of a group $G$ on a set $\Omega$ and its point stabilizers $G_{\alpha} = \{g \in G: \alpha^g = \alpha\}$ for $\alpha \in \Omega$. The action on each orbit is equivalent to the action of a point stabilizer on that orbit and we have that the point stabilizers on each orbit are conjugate and $G_{\alpha}^g = G_{\alpha^g}$, If $G$ acts transitive, this is key to the observation that for a point $\alpha$ the other points fixed by $G_{\alpha}$ are $\alpha^{N_G(G_{\alpha})}$ and by orbit-stabilizer we have $|N_G(G_{\alpha}) : G_{\alpha}|$ points that are fixed by $G_{\alpha}$ and each element that acts on the fixed point of $G_{\alpha}$ normalizes $G_{\alpha}$, hence conjugation and fixed points are closely related. Taking this further, if we have a single $g \in G_{\alpha}$, the total number of points fixed by $g$ is $|N_G(G_{\alpha}) : G_{\alpha}| \cdot k$, where $k$ denotes the number of distinct conjugates of $G_{\alpha}$ that contain $g$. If for example different point stabilizers intersect trivially (a situation that is not that uncommon), $\alpha^{N_G(G_{\alpha})}$ is precisely the set fixed by every single $g \in G_{\alpha}$. This is sometimes useful if you look at groups on which several restrictions on the fixed points of nontrivial elements are imposed.

Also as mentioned by Qiaochu Yuan two subgroups are conjugate iff the action on the cosets are equivalent. Further two transitive actions on two sets $\Omega, \Gamma$ of $G$ are equivalent if every point stabilizer of the action on $\Omega$ is also a point stabilizer of the action on $\Gamma$, and vice versa. Which is just another way to say this by the fundamental relation between conjugate point stabilizers and the points fixed by elements of $G$.

$\endgroup$
1
$\begingroup$

Isomorphic subgroups can be normal and not normal, which can be seen by taking the product of a group and a non-normal subgroup. So conjugacy can be seen in contrast as a reliable means of inferring the normality of many subgroups from one.

If you don't know representation theory as it's founded today, conjugacy of subgroups helps you imagine physical interpretations for the group, in that conjugacy of symmetry groups often replicates subgroups as a physically meaningful unit to other physically analogous units, such as in an affine space with its affine group, where conjugating by a translation takes the group of rotations about one point to another. This may still be relevant if you're looking for specific semantics, such as a Galois representation or representation by automorphisms of a simplicial complex.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.