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.


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    $\begingroup$ Your work seems fine. Here's a slightly unimportant comment: you do not need a new symbol $R$ to denote the conjugacy relation; you can use $\sim$ itself. Example usage: "$\sim$ is reflexive, because for all $a \in G$, we have $a = e a e^{-1}$." $\endgroup$
    – Srivatsan
    Jan 15, 2012 at 18:54
  • $\begingroup$ Thanks. I'm always looking for tips to clean up the style of my writeups. $\endgroup$
    – jobrien929
    Jan 15, 2012 at 19:07

2 Answers 2


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)?


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 $G$?


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?


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?


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.

  • $\begingroup$ Thanks @Srivatsan for the edit. $\endgroup$
    – user21436
    Jan 15, 2012 at 20:39
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    $\begingroup$ Thanks for all the time and thought you put into your response. It's a little beyond what the lectures have covered up to this point but you've given me a good preview of what's to come. I think what you said about group theory being more than just a collection of rules is true of every branch of mathematics. It seems to me that if you look at any area as just a bunch of theorems without seeing how they relate to each other than you're missing the point of what you're studying. $\endgroup$
    – jobrien929
    Jan 15, 2012 at 22:42
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    $\begingroup$ Having said that, understanding the rules well enough to be able to answer the problem sets is a prerequisite to gaining a fuller understanding of the subject. That's the reason I'm posting them here. To make sure I have a firm grasp on the building blocks needed to put the whole structure together. $\endgroup$
    – jobrien929
    Jan 15, 2012 at 22:47
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    $\begingroup$ @jobrien929 I am of the view that you misunderstand my intention. All I am telling you is you need to reflect on the exercises than merely solve them. So, if you are angry with me for telling you about exercise-solving, I am sorry. But, let me remind you, that the subject is intricate and subtle. $\endgroup$
    – user21436
    Jan 15, 2012 at 22:51
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    $\begingroup$ I'm not angry at all. Sorry if my response came across that way. Comments such as yours are exactly why I'm posting here. I'm interested in the conversations the answers start as well as the critiques of the answers themselves. I was merely pointing out that the point you made about group theory being subtle can also apply to math as a whole. I will accept an answer but I have been letting a day or two go by before doing so to see what answers I get. $\endgroup$
    – jobrien929
    Jan 15, 2012 at 23:02

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.

  • $\begingroup$ Thanks. I agree it's redundant. My intention was to first state my assumptions in the same terms used in the statement of the problem then to restate them in terms that would be useful for proving the statement. $\endgroup$
    – jobrien929
    Jan 15, 2012 at 19:12

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