Write the nonabelian group of order 21 as
replacing by gives the same group. Conjugation by splits the six nonidentity elements of the normal subgroup into the two classes
Each of the two other cosets is one conjugacy class:
Thus the class sizes are , so there are five irreducible characters.
The abelianization is , giving three linear characters. Put and , and define
The two nontrivial orbits of characters of the normal induce two irreducible characters of degree three. Their values vanish outside , and summing each orbit gives on the two nonidentity classes. The resulting character table is
The degree check
and character orthogonality relations verify that this is the complete table.

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