Could you explain how to show 7.18 in detail? I also included lists of definitions and theorems in the book as a reference. 7.14. Let G be a finite group with normal subgroup N. Show that there are at leastas many conjugacy classes in G as in G/N.7.18. Let G be a group and H the set of elements of G having only finitely manyconjugates. Show that H is a subgroup of G. Theorem 7.4. If G is any group, then conjugacy is an equivalence relation on theelements of G.Definition 7.3. Let G be a group and a e G. Then the conjugacy class of a is theset Ca= {g¬'ag : g e G}.-1Example 7.6. Note that Ca contains only a if and only if a e Z(G). (This happensif and only if g¬'ag = a for all g e G.)Example 7.7. Let k and n be positive integers with n > k. If G = S„ ando =(1 2 3 ... k), then we see from Lemma 6.2 that C, is the set of all k-cycles in G.It is important to know the size of a conjugacy class.Lemma 7.1. Let G be a finite group and a E G. Then the number of elements in Cais the index of the centralizer, [G : C (a)].Theorem 7.5 (Class Equation). Let G be a finite group, and let a1, ..representatives of the conjugacy classes in G with more than one element. Thenak be..• .|G| = |Z(G)|+[G: C(a1)]+ .+[G : C (ak)].Corollary 7.1. Let G be a group of order p", for some prime p and positive integern. Then the centre of G is not trivial.Corollary 7.2. Let G be a group of order p2, for some prime p. Then G is isomorphicto either Z,2 or Z, × Zp.Definition 7.4. Let G be a group and H a subgroup. We say that subgroups K andL of G are H-conjugate if there exists an h e H such that h-'Kh = L. WhenH = G, we simply say that K and L are conjugate.Example 7.8. Let G be S5 and H = ((1 3)(2 5 4)). Take o = (1 3)(2 4 5) =((1 3)(2 5 4))-l e H. Then we notice thato(1 23 4)o = (1 3)(2 5 4)(1 2 3 4)(1 3)(2 4 5) = (1 2 3 5).Therefore, for any integer i, o-(1 2 3 4)' o = (o¬1(1 2 3 4)0)' = (1 2 3 5)'. Thismeans that o-((1 2 3 4))o = ((1 2 3 5)). Thus, ((1 2 3 4)) and ((1 2 3 5)) areH-conjugate (and, therefore, conjugate).(o -1(1 2 3 4)o)Theorem 7.6. Let G be a group and H a subgroup of G. Then H-conjugacy is anequivalence relation on the set of all subgroups of G.Theorem 7.7. Let G be a finite group and H a subgroup. Then for any subgroup Kof G, the number of H-conjugates of K in G is [H : NH(K)].

Name: Could you explain how to show 7.18 in detail? I also included lists of
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Description: Could you explain how to show 7.18 in detail? I also included lists of definitions and theorems in the book as a reference. 7.14. Let G be a finite group with normal subgroup N. Show that there are at leastas many conjugacy classes in G as in G/N.7.1