$x$ conjugate to $y$ in a group $G$ is an equivalence relation on $G$ This is a question from the free Harvard online abstract algebra lectures.  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see this post.
This problem is from assignment 5.

a) Prove that the relation $x$ conjugate to $y$ in a group $G$ is an equivalence relation on $G$.
b) Describe the elements $a$ whose conjugacy class (= equivalence class) consists of the element $a$ alone.

a) Let $G$ be a group and $R$ be a relation on $G$ defined by $a\sim b$ if $a$ is conjugate to $b$.  Then $a\sim b$ if there is a $g\in G$ such that $a = gbg^{-1}$.  Let $a$ be an element of $G$.  Then $a=eae^{-1}$.  So $a\sim a$ and $R$ is reflexive.  Let $a$ and $b$ be elements of $G$ such that $a\sim b$.  Then there is a $g\in G$ such that $a=gbg^{-1}$.  Then $b=g^{-1}ag$.  Since $g^{-1}\in G$, $b\sim a$.  Hence, $R$ is symmetric.  Let $a,b$, and $c$ be elements of $G$ such that $a\sim b$ and $b\sim c$. Then there are elements $g,g^\prime\in G$ such that $a=gbg^{-1}$ and $b=g^\prime cg^{\prime -1}$.  Then $a=g(g^\prime c g^{\prime -1})g^{-1}=(gg^\prime)c(g^{\prime -1}g^{-1})=(gg^\prime)c(gg^\prime)^{-1}$. Since $gg^\prime\in G$, $a\sim c$.  Hence $R$ is transitive.  Therefore $R$ is an equivalence relation on $G$.
b) Let $S$ be the set of elements of $G$ such that, for $s\in S$, $s=gsg^{-1}$ for any $g\in G$.  Then $sg=gs$.  So $S$ is the set of elements that commute with every element of $G$.  In other words, $S$ is the center of $G$.
Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.
Thanks.
 A: I am writing this answer largely to push through that, Group theory is not a collection of discrete facts, but really a continuum of ideas about symmetry in Mathematics.
So, having solved this exercise, you can ask yourself the following questions.


*

*What is the number of equivalence classes (or since the relation is conjugation, conjugacy classes)?



Answer 
This may not have a nice answer for infinite groups. So, let's say we are interested only in finite groups, we make the following observations.
From (b), it is clear that the number of conjugacy classes is atleast the number of elements in the center of the group $Z(G)$. So, given the order, the group with least number of conjugacy classes are the abelian groups of that order. 
Now that you know that conjugation is an equivalence relation, and that equivalence relation partitions the set into disjoint sets,  you now have a new way looking at the order of the group.
Let $G$ be a group of finite order, and let $\{C_i\}_{i \in I}$ be the indexed collection of conjugacy classes. You now know that $C_i \cap C_j = \emptyset$ if $i \neq j$. So, you have 
$$\bigcup_{i \in I} C_i=G$$ 
This leads to a very important notion of class equation:
  $$|G|=\sum_{i \in I}C_i$$
As such this seems to be of little practical use. So, we'll improve this a little. From the way the orbits of a group action are defined, you will see that there is possibly a action whose orbits are exactly the equivalence classes here. With little thought, we can in fact see that this action must be the conjugation action of $G$ on $G$. So, you'll be able to prove the following facts from there:
  Since,
  $$|\mathscr{O}_x|=\dfrac{|G|}{|\operatorname{stab}~x|}$$ where $\mathscr {O_x}$ denotes the orbit of $x$ and $\operatorname{stab}~x$ denotes the stabilizer of $x$, you'll prove the following:
  $$|C_i|=\dfrac{|G|}{|C_G(x)|}$$ where $C_G(x)$ is the centralizer of $x \in C_i$. 
  Now this improves your class equation a little: (into the following form)
  $$|G|=|Z(G)|+\sum_{i\in I^*}\dfrac{|G|}{C_G(g_i)}$$ where $g_i \in C_i$ for $i \in I^*$ and $I^*$ indexes the conjugacy classes of cardinality greater than $1$.
Note that you also know the number of distinct conjugates of $g \in G$ from here. 



*

*What is so sacrosanct about having $~$ on elements of $G$? What would happen if $~$ were defined on $2^G$ where $2^G$ denotes the power set of $G$? 



Answer
Repeat the same kind of analysis that you did previously.



*

*Knowing that two elements of the same cycle type are conjugate in $S_n$, what does the above analysis mean to you? 



Answer
You must be able to prove that the number of conjugacy classes in $S_n$ equals the number of partitions of $n$ (and recollect Ramanujan!). Further, you can try your hand at getting the cardinality of conjugacy classes. 



*

*How do normal subgroups (whose definition resembles somewhat the relation we have defined!) look like in terms of conjugacy classes? 



Answer
You must be able to prove that it is a union of conjugacy classes. Further, you need to observe that, except when the normal subgroup is $\{e_G\}$, the normal subgroup is union of two or more conjugacy classes.

Please note that, I am completely sure that this doesn't answer your question. But, I write this answer to tell you that merely getting your solutions to the exercises right doesn't make you learn this subject. It is even more intricate and subtle.   
A: Your solution is okay. But I think one of your first two sentences (in (a)) is redundant: in fact you explain twice what ~ shall be. But apart from that everything is fine and straightforward.
