By Randall R. Holmes

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7, the function ϕ : G/N → G given by ϕ(aN ) = ϕ(a) is a well-defined homomorphism. By restricting the codomain to im ϕ we obtain an epimorphism G/N → im ϕ, which we continue to denote by ϕ. Let aN ∈ ker ϕ. Then ϕ(a) = ϕ(aN ) = e , so that a ∈ ker ϕ = N . Thus, aN = N . 6). Therefore, ϕ : G/N → im ϕ is an isomorphism and G/ ker ϕ = G/N ∼ = im ϕ. 2 Example: Classification of cyclic groups Let G be a cyclic group. Theorem. If G is infinite, then G ∼ = Z. Otherwise, G ∼ = Z/nZ, where n = |G|. Proof.

Let H, K ≤ G and assume that K ⊆ NG (H). (i) HK = KH, 37 (ii) HK ≤ G, (iii) H ∪ K = HK. In particular, if H G, then (i)–(iii) hold. Proof. (i) For each k ∈ K, we have Hk = kH k = kH. Thus HK = KH. (ii) First, e = ee ∈ HK. Next, using (i) we have (HK)(HK) = H(KH)K = H(HK)K = (HH)(KK) ⊆ HK, so HK is closed under multiplication. Finally, (HK)−1 = K −1 H −1 ⊆ KH = HK, so HK is closed under inversion. Thus, HK ≤ G. (iii) H = He ⊆ HK and K = eK ⊆ HK, so HK contains H ∪ K. Since HK is a subgroup of G (by (ii)) we have H ∪ K ⊆ HK.

If H ≤ K ≤ G, then H ≤ G, so the property of being a subgroup is transitive. However, it is not true in general that if H K G, then H G, as the following example shows. • Let G be the dihedral group D8 of order 8. 3. One easily checks that ρ2 , and hence N , is in the center of G so that N is normal. 5) and it contains H. 3), so K has order 4. 6. 3). This example shows that, in general, normality is not transitive. (Cf. ) 6 – Exercises 6–1 Give an example to show that it is possible to have a subgroup H of a group G satisfying both of the following: (a) There exists an element g of G such that g ∈ / NG (H) but H g ⊆ H.