Clarification on Symmetry Group My text says 
"It is a general fact, and an easy one to prove, that the invertible transformations of a mathematical object that preserve some feature of its structure always form a group.  We call this group the symmetry group of the object."
My question is why is this so?  How do we know that such transformations are associative?
 A: Invertible transformations are bijective functions and we know 
$$ (f \circ g) \circ  h = f \circ ( g \circ h ) $$ 
if $f,g,h$ are bijective.
A: Composition of functions or transformations is always associative. The proof is more or less just by unpacking the definition of function composition, so it can feel a little unsatisfying, but here it is:
Let $f,g,h$ all be functions/transformations mapping a set $X$ to itself. The statement that $(f\circ g)\circ h = f\circ (g\circ h)$ is an equality of functions, so to check it, we apply these functions to an arbitrary $x\in X$ and see if the answer is always the same:
$(f\circ g)\circ h (x)$ is $f\circ g$ applied to the result of applying $h$ to $x$. This is $f\circ g (h(x))$, which is $f(g(h(x)))$.
Meanwhile, $f\circ (g\circ h)(x)$ is $f$ applied to the result of applying $g\circ h$ to $x$. The latter is $g(h(x))$, so $f$ applied to this is $f(g(h(x)))$.
Note that the answers came out the same for arbitrary $x$, thus $(f\circ g)\circ h = f\circ (g\circ h)$.
Here is a "moral summary" of this proof: $(f\circ g) \circ h$ and $f\circ (g\circ h)$ both just mean "do $h$ then $g$ then $f$."
Aside: I think that the above becomes more interesting in light of the converse fact, that in any set with an associative operation, the set elements can be modeled by functions and the operation by function composition. I will make this precise in a moment, but the upshot is that everything associative is kind of like function composition. Function composition models something at the heart of what associativity means.
Anyway, here is my promised precise formulation. Let $(S,\star)$ be some set with some binary operation (i.e. a magma). Meanwhile, let $(F,\circ)$ be the magma consisting of the set of all functions from $S$ to itself, with the operation $\circ$ of function composition. By the above, $\circ$ is associative on $F$ (so $(F,\circ)$ is actually a semigroup). Now for any $x\in S$, let $\rho_x$ be the function $S\rightarrow S$ that describes $x$'s left action on $S$, i.e. for any $y\in S$, define $\rho_x(y)$ as $x\star y$. This gives us a map $\Phi$ from $S$ to $F$ sending $x$ to $\rho_x$.
I claim that the statement that $\star$ is associative (so $(S,\star)$ is really a semigroup) is equivalent to the statement that $\Phi: S\rightarrow F$ is a homomorphism of magmas (and thus a homomorphism of semigroups).
Proof: $\Phi$ is a homomorphism if and only if $\Phi(x\star y) = \Phi(x)\circ\Phi(y)$ for all $x,y\in S$. Since both sides are functions on $S$, we test equality by applying each side to all elements $z\in S$:
$$\Phi(x\star y)(z) = \rho_{x\star y}(z) = (x\star y) \star z$$
Meanwhile,
$$\Phi(x)\circ\Phi(y)(z) = \Phi(x)(\Phi(y)(z)) = \rho_x(\rho_y(z)) = \rho_x(y\star z) = x\star (y\star z)$$.
Thus $\Phi(x\star y)$ and $\Phi(x)\circ\Phi(y)$ are equal as functions for all $x,y$ if and only if $\star$ is associative for all $x,y,z$.
