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Answer to Question #23873 in Abstract Algebra for Hym@n B@ss

Question #23873
If a finite group G has at most three irreducible complex representations, show that G ∼ {1},Z2,Z3 or S3.
Expert's answer
Let r be the number of conjugacyclasses in G. By assumption, r ≤ 3. If r = 1, clearly G= {1}. Now assume r = 2. The two conjugacy classeshave cardinalities 1 and |G| − 1, so we have (|G| − 1) | |G|.This implies that |G| = 2, and hence G ∼ Z2. Finally, assume that r =3. Let 1, a, b be the cardinalities of the three conjugacy classes, andlet n = |G|. Then a + b = n − 1, and we mayassume that a ≥ (n − 1)/2. Write n = ac,where c ∈ Z. If c ≥ 3, then n ≥ 3a≥ 3(n − 1)/2 and hence n ≤ 3. In this case, G ∼ Z3. We may now assume c = 2,so b = n − a −1 = n / 2 − 1.
Since b|n, we can only have n= 4 or n = 6. If n = 4, G would have 4 conjugacyclasses. Therefore, n = 6, and since G is not abelian, G ∼ S3.

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