## Is there a symmetry group lurking behind every WLOG? [closed]

Most of us are introduced to "without loss of generality" before encountering formal group theory. To the uninitiated, the phrase almost seems like cheating, but soon we realize how intuitive and useful it is for simplifying and shortening proofs.

Perhaps this is a dumb question (in the sense that the answer might well be obvious), but is it true that behind every WLOG there is an implied symmetry group in play?

A Couple of Examples

(Schur's Inequality) If $a,b,c \ge 0$ and $r \ge 1$, then$$a^r(a-b)(a-c) + b^r(b-a)(b-c) + c^r(c-a)(c-b) \ge 0$$

Proof: Without loss of generality, assume $a \ge b \ge c$...
We can do this because the expression at hand is symmetric in $a,b,c$.
The group is $S_3$.

(Fundamental Theorem of Algebra) Every $n^{th}$-degree polynomial $a_n z^n + a_{n-1}z^{n-1} + \dots + a_1 z + a_0$ has a root in $\mathbb{C}$.

Proof: Without loss of generality, assume $a_n = 1$, because we can "divide through" by $a_n$...
The group is $\mathbb{R}- 0$ under multiplication.

-
I can imagine what a "no" answer to this question looks like, but I can't imagine what a "yes" answer would look like. – Qiaochu Yuan Jan 20 2012 at 1:47
Sometimes arguments in analysis begin "Let $\{x_n\}$ be a sequence which converges to $x$. Without loss of generality, assume $d(x,x_n) < 2^{-n}$..." Here the issue seems to be extra flexibility in the definitions rather than symmetry. – Paul Siegel Jan 20 2012 at 2:22
You might also want to allow groupoid symmetry since you might say without loss of generality assume the vertices of our graph is $\{1,\ldots,n\}$. – Benjamin Steinberg Jan 20 2012 at 3:49
You can attach a group action to any equivalence relation, but it is usually not canonical. – S. Carnahan Jan 20 2012 at 4:30