TITLE: What is an intuitive definition for "conjugate" in Group Theory?
QUESTION [34 upvotes]: 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)?

REPLY [4 votes]: 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.