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 ≥ *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.

Since

## Comments

## Leave a comment