This is too advanced for me. Not asking about proofs here.

Theorem 15.20: The set of all commutators $= \{aba^{-1}b^{-1} : a,b \in G \} $ generates $ \color{red}{\text{ and hence $\neq$} }$ (but can sometimes be) the commutator subgroup, call this $C$, of a group $G$. Then:
☼ $C \unlhd G$.
☼ If $N \unlhd G$, then $G/N$ Abelian $\iff C \le N$. This implies $G/C$ is Abelian.

p. 151: The commutators certainly generate a subgroup $C$.

Red color is mine. Fraleigh didn't define $C$ explicitly, but p. 2 of PDF says

$C = \{ x_1x_2...x_n : n \ge 1, \text{ each $x_i$ is a commutator in G } \}$ is the commutator subgroup.

(1.) The PDF proves $C$ is closed. But what if $n = 1$ in C?
Then $C = \{ x_1 : \text{ $x_1$ is a commutator in G } \} = \{aba^{-1}b^{-1} : a,b \in G \}$. I'm fretting about closure. Hence I take two elements and apply binary operation: $a_1b_1a_1^{-1}b_1^{-1} \circ a_2b_2a_2^{-1}b_2^{-1}$. Then?

(2.) What's the intuition for this theorem? Is this it? I'm still confounded.
How do you envisage and envision $C$ is the smallest $\unlhd G$ such that $G/G'$ commutes?

Fraleigh p. 150: Recall that in forming a factor group of G modulo [a normal subgroup N], we are essentially putting every element in G that is in N $ = e$, for N forms our new identity in the factor group. This indicates another use for factor groups...Thus we wish to attempt to form an abelianized version of G by replacing every commutator of G by $e$. By the first observation of this paragraph, we should then attempt to form the factor group $\color{blue}{\text{G modulo [the smallest normal subgroup we can find that $\subseteq$ all commutators of G]}}$.

(3.) I still don't understand why we are fretting about $\color{blue}{\text{G modulo [... commutators of G]}}$?

  • $\begingroup$ Your red "and hence $\neq$" is not necessarily true but usually true. In fact, the counter example of smallest order of a finite group in which the set of commutators doesn't equal the commutator subgroup is a group of order $\;96\;$ . $\endgroup$ – DonAntonio Feb 22 '14 at 9:39
  • $\begingroup$ The product of two commutators doesn't necessarily is a commutator, and this is why the set of comm's doesn't usually equal the comm. group. $\endgroup$ – DonAntonio Feb 22 '14 at 9:41
  • $\begingroup$ @DonAntonio: Thanks. I updated my question 1. $\endgroup$ – Group Theory Feb 23 '14 at 12:29

This is how I think of it:

Suppose $N$ is a subgroup containing every element of $G$ of the form: $aba^{-1}b^{-1}$ (these are called commutators).

If we knew $N$ was normal (back to that in a little bit) we could form the factor group $G/N$.

Now, in $G/N$ we have:

$(Na)(Nb)(Na)^{-1}(Nb)^{-1} = Naba^{-1}b^{-1} = N$

Which means that:

$(Na)(Nb) = (Nb)(Na)$, that is: $G/N$ is abelian.

OK, I promised I would talk about how we can be sure $N$ is normal. If we let $C$ be the set of all commutators, then the subgroup $\langle C\rangle$ of all finite products of commutators is certainly a subgroup containing all the commutators, and it's not hard to see it is the smallest such subgroup (we have to include at least all the commutators, and closure forces us to include all finite products).

Let's look at a conjugate of a commutator:

$g(aba^{-1}b^{-1})g^{-1} = ga(e)b(e)a^{-1}(e)b^{-1}(e)g^{-1}$

$= ga(g^{-1}g)b(g^{-1}g)a^{-1}(g^{-1}g)b^{-1}g^{-1}$

$= (gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})$

$= (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}$

In other words, a conjugate of a commutator is the commutator of the conjugates.

So the conjugate of any product of commutators is ALSO a product of commutators. This shows that $N = \langle C\rangle$ is normal.

A word about notation:

The expression $aba^{-1}b^{-1}$ is cumbersome to write, over and over again. The conjugates are also cumbersome to write over and over. So maths peoples have developed a "short-hand" by writing:

$aba^{-1}b^{-1} = [a,b]$

They (we?) also have a short-hand for writing:

$g^{-1}ag = a^g$ (with this notation we can say that: $(a^g)^h = a^{gh}$ and $(ab)^g = (a^g)(b^g)$, which is nifty).

In this shorthand, we can write:

$[a,b]^g = [a^g,b^g]$ which says what it took me 4 lines to write above (what I actually wrote was $[a,b]^{g^{-1}} = [a^{g^{-1}},b^{g^{-1}}]$ but I hope you get the picture).

With factor (quotient) groups, it is often common to suppress the "quotienting subgroup" and write:

$[a]$ or $\overline{a}$ instead of $Na$. In other words, we think of "factoring by $N$" as identically setting every element of $N$ to the identity (after all, $N = Ne = [e]$).

The "intuition" behind this is if $G$ were abelian, so that:

$ab = ba$


$aba^{-1}b^{-1} = e$ for every $a,b$.

If we "set" all these elements equal to $e$ (which is what factoring does), that makes $G$ abelian (we might have to squish it a little bit). We can actually say a bit more:

Suppose $\phi: G \to G'$ is a group homomorphism such that $\phi(G)$ is abelian. Then $[G,G]$ (this is another way our commutator subgroup is represented) is contained in the kernel of $\phi$. This says that the commutator subgroup is always the SMALLEST subgroup that "kills" (sends to the identity of $G'$) all commutators, which is called the universal property of the commutator subgroup.

Because this construction is done "in the same way" for any group $G$, it is said to be categorical, and is also called the Abelianization of $G$. The group $G/[G,G]$ is the group that is "most like" $G$ but turned into an abelian group. If $G$ is already abelian, then all the commutators are just $e$, so $G/[G,G] = G/\{e\} = G$.


1) it says $C$ is generated by elements in the form $aba^{-1}b^{-1}$. You do not need to write $aba^{-1}b^{-1}xyx^{-1}y^{-1}$ in the form of $(cdc^{-1}d^{-1})$. 'Generate' means you already take all possible multiplication; it is already an element of $C$.

2)First observe that $G/C$ is abelian in a very natural way.

Since $x^{-1}y^{-1}xy\in C\implies x^{-1}y^{-1}xyC=C\implies xyC=yxC \implies(xC)(yC)=(yC)(xC)$

Now let $N$ be a normal subgroup of $G$ s.t. $G/N$ is abelian.

Then, $xyN=yxN\implies x^{-1}y^{-1}xy\in N $ for all $x,y\in G$
But, if a group includes a set, then it also includes the group generated by that set.
Thus, we have $C\leq N$.

By the way, to show that $C$ is normal just observe that $[x,y]^z=[x^z,y^z]$.

If you understand first two answer, you will understand the third one.

I will explain the solution as Frank request.

Let $S\subset G$ then $<S>$ is set of all possile expression(or word if you like) created by elements of $S$.

Then suprisingly $<S>$ is a subgroup of $G$.(Why? if it is not obvious to you,you should study on subgroup generated by subset of $G$). One of the uniqe properties of $<S>$ is that it is the smallest subgroup of $G$ containing $S$.

Now,Let $S=\{aba^{-1}b^{-1}|a,b\in G\}$ then $<S>$ is ,of course, a subgroup.(as we said it is true for all subset of $G$).For $aba^{-1}b^{-1}xyx^{-1}y^{-1}$, it is an word which can be written by the elements of $S$,thus it is an element of $<S>$ .

$[x,y]$ is used to denote $xyx^{-1}y^{-1}$ and $g^{h}$ is used for $h^{-1}gh$.

Notice that; $[x^z,y^z]=z^{-1}xzz^{-1}yzz^{-1}x^{-1}zz^{-1}y^{-1}z=z^{-1}xyx^{-1}y^{-1}z=[x,y]^z$

Now,let $g\in <S>$ then it is a word, $$g=[a_1,b_1][a_2,b_2]..[a_k,b_k]$$ $$g^h=h^{-1}[a_1,b_1][a_2,b_2]..[a_k,b_k]h$$ $$=h^{-1}[a_1,b_1]hh^{-1}[a_2,b_2]hh^{-1}..h^{-1}[a_k,b_k]h$$ $$ =[a_1,b_1]^h[a_2,b_2]^h..[a_k,b_k]^h$$ $$=[a_1^h,b_1^h][a_2^h,b_2^h]..[a_k^h,b_k^h] $$ Thus,$g^h\in <S> \implies <S>$ is normal in $G$.

  • $\begingroup$ Thanks a lot. (1.) I know you don't need to, but how can you write $aba^{-1}b^{-1}(xyx^{-1}y^{-1})$ anyways? (2.) What do your square bracket mean for $[x,y]^z$? Can you please answer in your answer and not in comments? $\endgroup$ – Group Theory Feb 23 '14 at 12:29
  • $\begingroup$ @FrankMuer:I hope above answer is satisfactory for you. $\endgroup$ – mesel Feb 23 '14 at 17:11

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