# 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

Since

*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 ≥*3*a≥*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**∼**S*3.
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