The following is the incomplete operation table a 4-element group. The last…

Key insight: each row and each column of a group operation table must be a permutation of all group elements (Latin-square property).

• The identity e fixes every element, so the first row (and first column) is known: e, a, b, c.

• From the given entries in the row for a we read a*a = b, a*b = c, and a*c = e. Thus powers of a cycle as a, b, c, e (so a has order 4).

• View elements as powers of a: a = a^1, b = a^2, c = a^3, e = a^4. Compute products with c = a^3:

• c * e = a^3 * a^0 = a^3 = c

• c * a = a^3 * a^1 = a^4 = e

• c * b = a^3 * a^2 = a^5 = a

• c * c = a^3 * a^3 = a^6 = a^2 = b

Therefore the last row (the row labeled by c) is: c e a b.

Final check: every row and every column is a permutation of {e,a,b,c}, so the completed table is consistent with a group of order 4 (the cyclic group).