Unformatted text preview: ϕ : G → S 5 with kerϕ ≤ H .  H  = 2 implies ker ϕ = { e } or H . If kerϕ = H then H C G , which would be a contradiction. Thus kerϕ = { e } which implies ϕ is 1to1 and G ∼ = ϕ ( G ) ≤ S 5 . 6.  G  = 5 2 · 13 2 . n 5 ≡ 1 (mod 5) and n 5  13 2 . Thus n 5 = 1 and the Sylow 5subgroup, H , is normal. n 13 ≡ 1 (mod 13) and n 13  5 2 . Thus n 13 = 1 and the Sylow 13subgroup, K , is normal. H ∩ K = { e } since they are both pgroups for distinct p ’s. H,K C G and  HK  = 4225 =  G  thus HK = G and G ∼ = H × K . Since H and K are of order p 2 they are both abelian. Thus G is also abelian. Hence G ∼ = Z 25 × Z 169 , Z 5 × Z 5 × Z 169 , Z 25 × Z 13 × Z 13 , or Z 5 × Z 5 × Z 13 × Z 13 by the Fundamental Theorem of Finite Abelian Groups....
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 Spring '11
 Johnson
 Algebra, Normal subgroup, Abelian group, Finite Abelian Groups, 1Step Subgroup Test

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