Suppose that ˚is a homomorphism from a group Gonto Z 6 ⊕Z 2, and that the kernel of has order 5. Expalin why Gmust have normal subgroups of orders 5,10,15,20,30, and 60. Example Solution: Since Sker(˚)S=5, it follows form theorem 10.2.5 that ˚is 5 to 1. So SGS=Sker(˚)SSZ 6 ⊕Z 2S=5⋅6⋅2 =60. G is a normal subgroup of itself of order 60.
The cyclic group of order can be represented as (the integers mod under addition) or as generated by an abstract element . Mouse over a vertex of the lattice to see the order and index of the subgroup represented by that vertex; placing the cursor over an edge displays the index of the smaller subgroup in the larger subgroup.
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Proposition 5.8 A simple group of order 60 is isomorphic to A 5. Proof Let G be a simple group of order 60. The number of Sylow 5-subgroups of G is congruent to 1 (mod 5) and divides 12, but is not 1 (else the unique Sylow subgroup would be a normal subgroup of G). So there are six Sylow 5-subgroups.
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