It is not a group for the following reason.
The elements of P(X) are all the subsets of X.
A group must have an identity element, i.e. there is an subset e of X such that if Y is a subset of X then e U Y = Y U e = Y. Now since the empty set is a subset of every set and since Y U empty set = empty set U Y = Y for any Y then the empty set is the identity element.
But each element of a group has an inverse in the group, i.e. for any subset Y of X there is a subset Z of X such that Y U Z = Z U Y = the identity element = the empty set. Consider any non-empty subset of X, for example {a}. There is no subset Z of X such that {a} U Z = Z U {a} = the empty set since {a} U Z always contains at least a.
{P(X),U} is not a group since the non-empty subsets of X have no inverses.
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