Can a group have more than one operation? If not, what is the operation of a dihedral group? In my 1st year Mathematics BSc course, dihedral groups seems to include rotations and reflections, which suggests that groups can have more than one operation. But definitions of groups I've seen seem to suggest (not totally clearly) that groups have exactly one operation. 
If that's right, what's the single operation associated with dihedral groups? I would guess it's reflection, since rotations can be retrieved from compositions of reflections, but not the reverse.
 A: The operation associated with the dihedral group is combining transformations. For example, given an $n$-gon, I can first rotate it and then reflect it. Rotation is one member of the group, reflection - another, and their product in the group is the combined transformation.
A: Please be careful. In the context of groups (and more generally, algebras in the sense of General/Universal Algebra), an “operation” is something you do to elements of your set. So for a group $G$, an “operation” (or more specifically, a binary operation) for the group is usually understood to mean a function $G\times G\to G$, and more, it is understood to mean the function $G\times G\to G$ that has the properties that makes $G$ into a group. For the dihedral group, this operation is “composition”.
Instead, what you are talking about are the ways the group acts on something (for the dihedral group, the way it acts on a regular $n$-gon); the elements of the dihedral group have two different types of actions on the regular $n$-gon, the rotations (which preserve orientation) and the reflections (which reverse orientation).

Note that while groups are usually defined with a single binary operation, as follows:

A group is a set $G$ together with a binary operation $\cdot\colon G\times G\to G$, usually written in infix notation, such that:
  
  
*
  
*For all $a,b,c\in G$, $(a\cdot b)\cdot c = a\cdot (b\cdot c)$ ($\cdot$ is associative);
  
*There exists $e\in G$ such that for all $a\in G$, $e\cdot a = a\cdot e = a$ ($\cdot$ has an identity element);
  
*For every $a\in G$ there exists $b\in G$ such that $a\cdot b=b\cdot a = e$ (every element has a $\cdot$-inverse).

However, from the viewpoint of Universal Algebra, groups are actually sets with three operations: a binary, a unary, and nullary operation (two, one, and zero inputs). As follows:

A group is an ordered pair $(G,\Omega)$, where $\Omega=\{\cdot,{}^{-1},e\}$ consists of a binary operation $\cdot$ on $G$, a unary operation ${}^{-1}$ on $G$, and a nullary operation $e$ on $G$, such that:
  
  
*
  
*$\cdot$ is associative;
  
*$a\cdot e = e\cdot a = a$ for all $a\in G$;
  
*$a\cdot a^{-1}=a^{-1}\cdot a= e$ for all $a\in G$.
  

A: A group $(G,\star)$ has only one operation, which is the binary operation $\star$ between its elements. If a group $G$ contains bijective maps from some fixed set $X$ to itself (such as geometric transformations of the plane are, or the permutations of a set of $n$ elements) then the operation of the group is the composition of functions $f\circ g$, where $f\circ g$ is the function mapping $x\mapsto f(g(x))$.
A: Well, with all the good answers by Arturo and Gae, there is still another angle at this for finite groups. Given a finite group $G$ with the binary operation $\cdot$, one can construct, using that operation, another operation on the same underlying set. What I am alluding to is the so-called Baer Trick: suppose that for the finite group $(G,\cdot)$, $G/Z(G)$ is abelian and $|G|$ is odd. Then there exists an  operation $x \star y$, defined for elements $x,y \in G$, that turns $(G,\star)$ into an abelian group! In this construction, surprisingly, the order of an element in $(G,\cdot)$ equals its order in $(G,\star)$. The Baer Trick is used for example in the theory of coprime actions of groups (see also Kurzweil-Stellmacher, The Theory of Finite Groups, An Introduction, Chapter 8.5)
