Can an element other than the neutral element be its own inverse? Take the following operation $*$ on the set $\{a, b\}$:


*

*$a * b = a$

*$b * a = a$

*$a * a = b$

*$b * b = b$


$b$ is the neutral element. Can $a$ also be its own inverse, even though it's not the neutral element? Or does the inverse property require that only the neutral element may be its own inverse but all other elements must have another element be the inverse.
 A: Your set is isomorphic to the two-element group: $b=1$, $a=-1$, $*=$multiplication. So yes, $a$ can very well be its own inverse.
A: Yes, an element other than the identity can be its own inverse. A simple example is the numbers $0,1,2,3$ under addition modulo 4, where 0 is the identity, and 2 is its own inverse. 
A: Denoting 1 for the identity element, we have for every group element $a$:
$a=a^{-1}\Leftrightarrow a^2=1$.
So a non-identity element is its own inverse iff it has order 2. This is perfectly possible, as Gerry Myerson showed. Or look at the group of automorphisms of $\mathbb{C}-\{0\}$ (invertible ring homomorphisms), the identity element being the identity map, and group multiplication being composition of automorpisms. Then complex conjugation has order 2: it sends $x+iy$ to $x-iy$, and the latter is sent back to $x+iy$.
A: All the elements that are its own inverses constitute a special example of the so-called elements of finite orders; in the finitely generated commutative groups, these elements constitute a subgroup, called the torsion group, of finite order itself.  Of course then, these elements need not be neutral elements.
A: First off, let's get some definitions straight.
Definition 1: A neutral n on a set S with a binary operation * we define as an element n of S such that for all s belonging to S, (s*n)=s and (n*s)=s.
Definition 2: An inverse i on a set S of an element j of S which has neutral n and binary operation * we define as an element i of S such that (j*i)=n and (i*j)=n.  j and i need not come as distinct, texts just use different letters so that they may come as distinct, as they often do.
Now for the operation * on {a, b} described in the original post, we have b as the neutral.  We have (a*a)=b also.  Thus, by definition 2 element "a" of {a, b} qualifies as its own inverse.  So, you have a simple example in the original post.
