if A = {a,b,c} . Prove OR disprove that P(A) "power set of A" is a group under the union operation So my thinking is that 

$$P(A) = \{\{\emptyset\},\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\} $$

although it's clearly closed under union , assiociatve and the identity exists which is the empty set but still it's not a group since there are no inverses exists for any of the elements except for that of the empty set itself.

But is that valid to say , I mean how can I do it more rigorously?
The empty set is the identity but $B\cup C$ contains $B$ so if $B$ is non empty $B\cup C$ is non empty.  Hence a nonempty element is not invertible.
Let $~ X= \{1,2,3\}~$ and $~P(X)~$ be the power set of $~X~$. Examine if $~P(X)~$ is a group under the composition $~*~$ defined by $$~A*B= A\cup B ,~~~~~~~ \forall A,B \in P(X)~.$$