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The Dihedral group makes sense, "Di" means two, and "hedral" means.. shape I think (I've just realised how much of what I think words mean are guesses based on experience) like a "polygon" is a 2d shape, a "polyhedron" I've always thought is just "any shape" (example 3 dimensions). So "Dihedral" means 2 dimensions, which it is. (Octagon and cube have different ... reflection+rotation groups (I hope))

The symmetric group has nothing to do with symmetries of shapes, it instead is the group of all bijections from one set to another. I suppose this means symmetry as in "abc->acb" has the "symmetric" version "$abc\to bac$".

Now the alternating group... I have no idea.

Now I discovered these in "Algebra" and "Abstract Algebra" but I've peeked at the contents page of the Combinatorics book I am reading and the symmetric is discussed there, this makes a lot of sense really, combinatorics seems to be the study of finite functions and seems to be easier than functions on the reals and stuff (which felt odd as I was sketching $x^3$ before I was considering "discreet" functions, but if anything they're easier)

So I guess that they were named because Combinatorics got there first. How? What (general case) question was being faced that lead to the solution being called "The symmetric group" (or even set, I doubt combinatorics would care that it is a group).

Sometimes how and why are similar, like turning points, "how did we decide to call $\frac{dy}{dx}=0$ a turning point" and "why" are identical. However there may be many "whys" with these groups as they apply in many areas, there can be only one how. Where the term first cropped up would be a good starting point, but I've yet to find anything.

I'll answer this myself if my searching yields anything.

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Just a guess: the map $S_n\to \pm 1$ alternates between $1$ and $-1$, and has kernel $A_n$, which may be the source of that name. – Alex Becker Mar 26 '14 at 1:48
@AlexBecker I hate to ask this (I may look daft) but what is that map? The map that assigns a permutation on n to + or -1? (how would it do it), I'm also not sure what a kernel would be in that case, unless 1 is the identity permutation. – Alec Teal Mar 26 '14 at 1:50
There are various ways to define it. The simplest is probably to write a permutation as a matrix and take its determinant. $\pm 1$ is just the group $\{1,-1\}$ under multiplication, so $1$ is the identity and the kernel is the preimage of $1$. – Alex Becker Mar 26 '14 at 1:58
I don;t think there's any reason to feel silly; everyone had to learn that at some point. But feel free to delete the comment if you wish. – Alex Becker Mar 26 '14 at 2:02

This is taken from Timothy Chow's answer at, which addresses your question:

In Burnside's book, Theory of Groups of Finite Order, when defining the symmetric and alternating groups, he puts an asterisk next to the word "alternating" and adds the following footnote:

"The symmetric group has been so called because the only functions of the $n$ symbols which are unaltered by all the substitutions of the group are the symmetric functions. All the substitutions of the alternating group leave the square root of the discriminant unaltered."

By "the square root of the discriminant," Burnside means the polynomial $$\prod_{r=1}^{n-1}\prod_{s=r+1}^n (a_r-a_s),$$ which of course is an alternating polynomial.

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I did a small edit to add the name of the author of the answer you are quoting. – Andrés E. Caicedo Mar 26 '14 at 1:59
@AndresCaicedo Thanks for the helpful edit! – Sanath K. Devalapurkar Mar 26 '14 at 2:00
@ sanath: good reference – Groups Jun 7 '15 at 5:09

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