Answer:
The answer is the proof so it is long.
The question doesn't define u(n), but it's not hard to guess.
Group G with operation ∘
For all a and b and c in G:
1) identity: e ∈ G, e∘a = a∘e = a,
2) inverse: a' ∈ G, a∘a' = a'∘a = e,
3) closed: a∘b ∈ G,
4) associative: (a∘b)∘c = a∘(b∘c),
5) (optional) commutative: a∘b = b∘a.
Define group u(n) for n prime is the set of integers 0 < i < n with operation multiplication modulo n.
If n isn't prime, we exclude from the group all integers which share factors with n.
Identity: e = 1. Clearly 1∘a = a∘1 = a. (a is already < n).
Closed: u(n) is closed for n prime. We must show that for all a, b ∈ u(n), the integer product ab is not divisible by n, so that ab ≢ 0 (mod n). Since n is prime, ab ≠ n. Since a < n, b < n, no factors of ab can equal prime n. (If n isn't prime, we already excluded from u(n) all integers sharing factors with n).
Inverse: for all a ∈ u(n), there is a' ∈ u(n) with a∘a' = 1. To find a', we apply Euclid's algorithm and write 1 as a linear combination of n and a. The coefficient of a is a' < n.
Associative and Commutative:
(a∘b)∘c = a∘(b∘c) because (ab)c = a(bc)
a∘b = b∘a because ab = ba.
.</h2>