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.

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closed as not a real question by Yemon Choi, Benjamin Steinberg, Mariano Suárez-Alvarez♦, Alain Valette, Bill JohnsonJan 20 '12 at 17:11

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

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 '12 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 '12 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 '12 at 3:49
You can attach a group action to any equivalence relation, but it is usually not canonical. –  S. Carnahan Jan 20 '12 at 4:30

Fix a semi-formal language for discourse and proof. (I will let you worry about niggling issues such as consistency and soundness.) Now consider proving your statement with as much detail and completeness as your semi-formal system allows. Your proof may break into subcases and subproofs, and you may find it convenient to refactor your proof so that you make repeated use of some lemmas. At some point you may have a situation like wanting to establish in some contexts two statements, say phi(a,b) and phi(b,a), and you find that they have similar proofs. At this point, you can define an equivalence relation between tuples in the language, and you can construct and investigate symmetries of the syntactic elements of the proof as represented in the semi-formal system. So you can arrange things so that the answer to your question is yes. Whether it is prudent to take such relations from the semi-formal system back to the realm of your objects of study is likely very situation-specific.