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In Abstract Algebra, I learned about "conjugation" in the context of a group $H$ being a 'normal' subgroup of $G$ if the element $xhx^{-1}\in H$ for any $x\in G$. But this is not the first time I've seen the word 'conjugate'. The other times I've seen this are in pre-calculus, when trying to rationalize a denominator, or in the case where $(x+y)$ is the conjugate of $(x-y)$. Does the Group Theory version of conjugate have any link to the pre-calculus version (and other uses)?

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    $\begingroup$ This is a good question with the answer a bit "yes" and "no". The answer involves Galois Theory, where a "conjugate" in the "pre-calculus" situations is in fact an image of an element under the action of a group - and is generalised. But that is not the same thing as conjugation within the group. I am sure someone will be able to give a fuller and clearer explanation, which is why this is a comment rather than an answer. $\endgroup$ – Mark Bennet Dec 11 '15 at 19:35
  • $\begingroup$ A conjugate element, in each of these contexts, is one that "acts like" the original. In your cases, the element $gxg^{-1}$ acts like $x$, and $a+\sqrt b$ acts like $a-\sqrt b$. $\endgroup$ – Omnomnomnom Dec 11 '15 at 20:07
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C. Falcon's answer refers to Galois theory, which is almost certainly the historically correct answer. But the connection between the two ideas can be explained in slightly more elementary terms:

If $G$ is a group acting on a set $S$, and $s,t\in S$ lie in the same orbit, then the stabilizers of $s$ and $t$ are conjugate subgroups of $G$. Specifically, if $t = hs$, then $gs = s \iff (hgh^{-1})t = t$.

Since we frequently study group elements in terms of their fixed points, it makes sense that the terminology for conjugation in a group gets sometimes mixed up with the terminology for orbits.

For example, we can check that $a+b\sqrt{2} \mapsto a-b\sqrt{2}$ defines an action of the cyclic group of order $2$ on the field $\mathbb{Q}(\sqrt{2})$. Then $a+b\sqrt{2}$ is conjugate to $a-b\sqrt{2}$ in the sense that they lie in the same orbit under this action.

More generally, Galois theory tells us that if $p(X)$ is any irreducible polynomial over a field $k$, and $k\subset K$ is a splitting field, then there is a group acting on $K$ that fixes $k$ and acts transitively on the roots of $p$. In other words, the roots of $p$ are conjugate.

This is somewhat of an abuse of terminology, but it may explain the conceptual link between the two different uses of the word.

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Arguably the easiest and the most intuitive way to understand conjugation is through linear algebra.

$T : V \to V$ be a linear transformation of the vector space $V$. Then given a basis $\Bbb B = \{e_1, \cdots, e_n\}$ for $V$, one can represent $T$ by a matrix $[T]_{\Bbb B} = [T(e_1), \cdots, T(e_n)]$, with each $T(e_i)$ expressed in terms of $\Bbb B$. That is, the columns of the matrix are the images of each element of the basis $\Bbb B$ under the transformation.

If $\Bbb B = \{v_1, \cdots, v_n\}$ and $\Bbb B' = \{w_1, \cdots, w_n\}$ are two bases of $V$, then we can try to see what happens if we start with the basis $\Bbb B$ and then switch to the basis $\Bbb B'$ by expressing $v_i$ as linear combination $\sum_k a_{ki} w_i$ of elements of $\Bbb B$. Similarly, we can try expressing elements $w_i$ of $\Bbb B'$ as linear combinations $\sum_k a_{ki}' v_i$ of elements of $\Bbb B$.

The matrix $P = [a_{ki}]$ is called the basechange matrix obtained from changing base from $\Bbb B$ to $\Bbb B'$. It can be easily verified that $[a'_{ki}]$ is simply $P^{-1}$.

The relevant fact is $[T]_{\Bbb B'} = P^{-1} [T]_{\Bbb B} P$. Thus, conjugation of matrices correspond to change of basis and conjugacy classes of matrices in $GL(V)$ correspond to basis-free linear operators.


This description of conjugacy can even be seen in general groups. $G$ be an arbitrary group which acts on a set $X$ by

$$G \times X \to X \\ (g, x) \mapsto gx$$

If $g$ acts on elements on $X$ by multiplication, $hgh^{-1}$ does the same job $g$ does, but on $hx$ instead of $x$, and then puts everything back at place. One can easily see this by letting $\text{Isom}(\Bbb R^3)$ act on $\Bbb R^3$ by isometries, say, and letting $g$ and $h$ to be elements of the isometry group (e.g., rotation/translation).

Thus, $hgh^{-1}$ is not really very different from $g$, but it just does the work of $g$ at $hx$ instead of $x$. This is more general in the sense that letting $G = GL(V)$ and $G$ act on the vector space $X = V$ gives me back the linear algebra picture.


To answer the main question, to my mind, Galois conjugates are not really related to conjugates as in groups (which has already been pointed out by several answers here). The real motivation for conjugation comes from Galois theory in a different way.

Namely, if $\text{Gal}(L/K)$ is the group of $K$-automorphisms of $L$, then given an intermediate extension $L/E/K$ (terminologies are explaned in C. Falcon's answer), $\text{Gal}(L/E)$ is naturally a subgroup of $\text{Gal}(L/K)$, as $E$-automorphisms of $L$ are automatically $K$-automorphisms of $L$ (pointwise fixing something larger automatically pointwise fixes the subset).

Now, if $L/E/K$ is a tower of Galois extensions, then $\text{Gal}(L/E)$ is normal in $\text{Gal}(L/K)$. This is where the notion of conjugacy comes to play, because Galois wanted to define group structure on the collection of all cosets.

If $G$ is a group, $N \leq G$ is a subgroup, it's easy to see where conjugacy comes in trying to define group structure on $G/N$. If $g_1N, g_2N$ are two elements in $G/N$, then the most obvious way to try to define group structure is as $(g_1 N)\cdot(g_2 N) = (g_1g_2)N$. This can be made well-defined if and only if $gNg^{-1} = N$ (one direction is easy, for the other direction if $gN = g'N$, pick $h \in N$. $hN = eN$ implies $(hN) \cdot (gN) = gN$, which is - as we claim - the same as $(hN) \cdot (gN) = (hg)N$. Hence $(g^{-1}hg)N = N$, and $N$ is normal, as desired)

It's interesting as a remark that Galois deduced that the quotient $\text{Gal}(L/K)/\text{Gal}(L/E)$ is in fact isomorphic to $\text{Gal}(K/E)$, although that's irrelevant here.

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According to A. Jeanneret and D. Lines' book Invitation à l'algèbre the group conjugation arises from Galois' theory. Let me introduce you to the basic concepts.

Definition. Let $K$ be a field. If $L$ is a field and $\varphi:K\hookrightarrow L$ is a ring homomorphism $(L,\varphi)$ is said to be a field extension of $K$ and one writes $L/K$.

Remark. A ring homomorphism $\varphi$ from a field $K$ to a field $L$ is always injective. Indeed, let $x\in K$ such that $\varphi(x)=0_L$ and assume by contradiction $x\neq 0_K$, $x\in K^\times$. Since a ring homomorphism is unitary, one has: $$1_L=\varphi(xx^{-1})=\varphi(x)\varphi(x)^{-1}=0_L,$$ a contradiction, since a field is not the null ring. Hence, $x=0_K$ and $\varphi$ is injective.

Example. $\mathbb{R}/\mathbb{Q}$ is a field-extension for the natural inclusion.

Definition. Let $L/K$ be a field extension, $M$ is said to be an intermediate field if and only if $L/M$ and $M/K$ are field extensions.

Example. $\mathbb{R}$ is an intermediate field of $\mathbb{C}/\mathbb{Q}$.

Proposition-Definition. Let $L/K$ be a field extension and $M$ an intermediate field. If $\sigma\in\textrm{Aut}(L)$ (ring isomorphisms from $L$ to $L$), $\sigma(M)$ is an intermediate field called the conjugate of $M$.

Remark. The complex conjugation is a ring automorphism of $\mathbb{C}$. The field conjugation defined above is thus a generalization of the complex conjugation.

Proposition-Definition. Let $L/K$ be a field extension, $\textrm{Gal}(L/K):=\textrm{Aut}_{K}(L)$ (ring isomorphisms from $L$ to $L$ whose restrictions to $K$ is identity) is a group for the composition called the Galois' group of the extension $L/K$.

Proof. Let us show that $\textrm{Gal}(L/K)$ is a subgroup of $\textrm{Aut}(L)$.

  • $\textrm{id}_L\in\textrm{Gal}(L/K)$.

  • Let $\sigma,\tau\in\textrm{Gal}(L/K),\sigma\circ\tau\in\textrm{Aut}(L)$. Let $x\in K$, one has: $$\sigma\circ\tau(x)=\sigma(x)=x.$$ Hence, $\sigma\circ\tau\in\textrm{Gal}(L/K)$.

  • Let $\sigma\in\textrm{Gal}(L/K)$, $\sigma^{-1}\in\textrm{Aut}(L)$. Let $x\in K$, one has: $\sigma(x)=x$, hence one gets: $$x=\sigma^{-1}(x).$$ $\sigma^{-1}\in\textrm{Gal}(L/K)$.

Whence the result. $\Box$

The link with the group conjugation is contained in the:

Proposition. Let $L/K$ be a field extension, $M$ be an intermediate field and $\tau\in\textrm{Gal}(L/K)$, one has: $$\textrm{Gal}(L/\tau(M))=\tau\textrm{Gal}(L/M)\tau^{-1}.$$

Proof.

  • Let $\sigma\in\textrm{Gal}(L/K)$ and let $y\in\tau(M)$, there exists $x\in M$ such that $y=\tau(x)$, one has: $$\tau\circ\sigma\circ\tau^{-1}(y)=\tau\circ\sigma(x)=\tau(x)=y.$$ Hence, $\tau\circ\sigma\circ\tau^{-1}\in\textrm{Gal}(L/\tau(M))$ and one has: $$\tau\textrm{Gal}(L/K)\tau^{-1}\subseteq\textrm{Gal}(L/\tau(M)).$$

  • Let $\rho\in\textrm{Gal}(L/\tau(M))$ and let define $\sigma:=\tau^{-1}\circ\rho\circ\tau$. Let $x\in K$, one has: $$\sigma(x)=\tau^{-1}\circ\rho\circ\tau(x)=\tau^{-1}\circ\tau(x)=x.$$ Hence, $\sigma\in\textrm{Gal}(L/M)$ and since $\rho=\tau\circ\sigma\circ\tau^{-1}$, $\rho\in\tau\textrm{Gal}(L/M)\tau^{-1}$. Therefore: $$\textrm{Gal}(L/\tau(M))\subseteq\tau\textrm{Gal}(L/K)\tau^{-1}.$$

Whence the result. $\Box$

Remark. In other words, this proposition states that the Galois' group associate with a conjugate field of $M$ is a group conjugate of the Galois' group associate with $M$.

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    $\begingroup$ It's clear from your work, but perhaps you want to emphasize that your ring homomorphisms are always unitary. (Some people think of the zero map as a ring homomorphism.) $\endgroup$ – Max Dec 11 '15 at 19:55
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The notion of conjugacy was first developped by Cauchy for permutation groups. For permutation groups it's extremely simple to understand, and in fact since all groups are isomorphic to some permutation group, you can take this as a general understanding.

Say I've got $10$ different objects which I'm swapping about. I can consider two permutations: one in which I swap the first two, and another in which I swap the last two. But these permutations are really the same thing, right? They're the same permutation, just done on different objects - I perform the same "action" in both cases, if you see what I mean. For a more complex case, imagine if I take the first five objects and rotate their positions, then swap the first and fifth. This is "the same" as if I took the last five objects and performed the same action on them (the mere fact that you're able to understand what I mean by "perform the same action on them" without my having to explain it in detail is proof of this).

Two permutations are conjugate if they are "the same permutation done to different objects" in this sense. How can we make this more precise? Let's work out an example: the permutations $P = (1\ 2)$ and $Q = (3\ 4)$, on say the set of numbers between $1$ and $10$. The second one is just $P$, but done to $3$ and $4$ rather than $1$ and $2$. So consider the permutation $S$ which sends $4$ to $2$ and $3$ to $1$. We have: $$P=S^{-1}QS$$ We first of all apply $S$ - this is basically just "renaming" the elements of our set. Then we apply $P$, and then we undo $S$ to get back to our original "naming scheme". It's admittedly a bit hard to explain this in writing, but hopefully with some thinking you should be able to see that the existence of an $S$ such that $P=S^{-1}QS$ really does capture the idea that $P$ and $Q$ are "the same thing done to different objects".

By the way, for finite permutation groups every permutation can be written as a product of disjoint cycles, and conjugacy is equivalent to having the same "structure" in one's cycle decomposition. Defining exactly what I mean by "structure" is one of those things that's hard to explain in words but easy to see with an example. The permutations $(3\ 4)(5\ 6\ 8)$ and $(2\ 3)(1\ 4\ 9)$ are clearly "the same", but done to different objects. The fact that both are the product of a 2-cycle and a disjoint 3-cycle suffices to show that they're conjugate.

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Conjugation usually relates to mappings on the algebraic structure called automorphisms. These are bijective maps which preserve the operation of the structure.

In $\mathbb{C}$ there are two operations, addition and multiplication. An automorphism of $\mathbb{C}$ is then a bijection from $\mathbb{C} \to \mathbb{C}$ which preserves both of these options. Since $\mathrm{i}$ and $-\mathrm{i}$ are both solutions to $x^{2} +1 = 0$, it can be shown that the map $a+b\rm{i} \mapsto a-b\rm{i}$ provides an automorphism of $\mathbb{C}$. So if you have $z$ and $\bar{z}$ equivalent under this map, they are called conjugates. Notice that if you multiply two complex conjugates together you get $(a + b\rm{i})(a-b\rm{i}) = a^{2}+b^{2} \in \mathbb{R}$, this relates to the case in precalculus where you are rationalizing a denominator by multiplying $(a+b\sqrt{d})(a-b\sqrt{d}) \in \mathbb{Q}$; in this setting, there is also a connection to field automorphisms of $\mathbb{Q}(\sqrt{d})$.

In a group $G$ there is only one operation. One common example of an automorphism of a group is conjugation by an element $g \in G$, this is a bijective map from $G \to G$ given by $x \mapsto g^{-1} x g$. Two elements $h_{1}, h_{2} \in G$ are called conjugates if there exists some $g \in G$ such that $h_{2} = g^{-1} h_{1} g$. (Note that there are sometimes other automorphisms of a group that are not equivalent to conjugation by an element. Group elements that are equivalent under these other types of automorphisms are usually not called conjugates).

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