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.