Abstract Algebra: Theory and Applications · Problem 15.4.5

Rings — Thomas Judson

Mathematics — Abstract Algebra · sylow-theorems

Problem

Prove that no group of order 96 is simple. Hint. Since

|G| = 96 = 2^5 \cdot 3

G

has either one or three Sylow 2-subgroups by the Third Sylow Theorem. If there is only one subgroup, we are done. If there are three Sylow 2-subgroups, let

H

and

K

be two of them. Therefore,

|H \cap K| \geq 16

; otherwise,

HK

would have

(32 \cdot 32)/8 = 128

elements, which is impossible. Thus,

H \cap K

is normal in both

H

and

K

since it has index 2 in both groups.

Related theory

Sylow S Theorems

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