In a document, answer the following questions:

Solution 1:
By definition of conjugacy, two elements a and b from group G are called conjugate if there is an
element g ϵ G such that gag-1 = b.
Let us denote this conjugacy relation by R.
Therefore, aRb iff gag-1 = b.
(i) Reflexivity Of R:
a = 1.a (where 1 is identity element of G)
=> a = 1.a.1-1 (Because 1-1 = 1)
=> aRa
=> R is reflexive relation.
(ii) Symmetry Of R:
Now if aRb => There is an element g ϵ G such that gag-1 = b
=> (gag-1) g = bg
=> (ga)(g-1g) = bg
=> ga = bg (Because g-1 g = 1, the identity element of G)
=> g-1 (ga) = g-1bg
=> (g-1g)a = g-1bg
=> a = g-1b (g-1)-1 (Because (g-1)-1 = g) - (1)
Let g-1 = h ϵ G.
From (1) we can say that there is an element h ϵ G such that hbh-1 = a i.e. bRa. From this we get
that R is symmetric.
(iii) Transitivity Of R:
Let a,b,c are elements of G such that aRb and bRc.
=> there is g ϵ G and h ϵ G such that a = gbg-1 and b = hch-1
=> a = g(hch-1)g-1
=> a = (gh)(c)(h-1g-1)
=> a = (gh)(c)(gh)-1 (Because (gh)-1 = h-1g-1)
=> aRc
=> R is transitive
From above discussion we get R is an equivalence relation on G i.e conjugacy is an equivalence relation.
Solution 2:
(i) To show Cg = {g} => g ϵ Z(G):
Cg = {aga
-1 : a ϵ G} (By definition of Cg)
=> {g} = {aga-1 : a ϵ G}
=> aga-1 = g for all a ϵ G.
=> ag = ga for all a ϵ G.
=> g ϵ Z(G) (Because Z(G) = {g...