Show that $[G,G]$ is a normal subgroup of $G.$

Question

Here is the question I want to answer:

In a group $G,$ the commutator of $x,y \in G$ is $[x,y] = xyx^{-1}y^{-1}.$ Let $[G,G]$ be the subgroup generated by all commutators in $G,$ noting that if $G$ is abelian, then $[G,G] = 1.$\ (a) Show that $[G,G]$ is a normal subgroup of $G.$

Here was my answer:

To prove this we need two identities:

$$[x,z]^{-1} = [z,x]. \tag{*}$$

Proof:

Follows directly from the definition of the commutator of $x$ and $y.$ As $[x,y] = xyx^{-1}y^{-1},$ then $[x,y]^{-1} = yxy^{-1}x^{-1}$ which is the commutator of $y,x$.

$$[x,yz] = [x,z][x,y]^z. \tag{**}$$

Proof:

Since the left-hand side is: $$[x,yz] = xyzx^{-1}y^{-1}z^{-1} \tag{1}$$ and since $[x,y]^z = z [x,y] z^{-1},$ then the right-hand side is: $$[x,z][x,y]^z =(xzx^{-1}z^{-1})(z (xyx^{-1}y^{-1}) z^{-1}) = xzyx^{-1}y^{-1}z^{-1} \tag{2}$$

Therefore, from $(1)$ and $(2)$ it is clear that the second identity is correct.

Showing that $[G,G]$ is a normal subgroup of $G.$

We want to show that $\forall a \in [G,G], \forall z \in G,$ we have $zaz^{-1} \in [G,G]$ where $a = [x,y].$

Proof:

Since we have that $[x,yz] = [x,z][x,y]^z$ from the first identity that I proved at the beginning, then multiplying it by $[x,z]^{-1}$ from the left, we get$$[x,z]^{-1}[x,yz] = [x,y]^z.$$ Which using $[x,z]^{-1} = [z,x],$ is equal to $$[z,x][x,yz] = [x,y]^z.$$But we know that $[G,G]$ is a subgroup by the given and hence a group and so it satisfies the closure property i.e. $[z,x][x,yz] \in [G,G]$ So $[x,y]^z \in [G,G]$ as required.

My question is:

I was told that my answer is incomplete and that the question is asking to prove that $[G,G]$ is a normal subgroup of $G$ and not the commutator $[x,y].$ Could anyone explain more to me that and how can I complete my answer please ?

Thanks in advance.

Answer 1

Let $x\in[G,G]$ and let $g\in G$. Then $gxg^{-1}=[g,x]x$ is also in $[G,G]$

Answer 2

The problem is that, generally, not every element of the commutator is a commutator.

To rectify this, you need to show that "reduced words" formed out of commutators are conjugated into reduced words formed out of commutators.

You can do that by showing that commutators are conjugated into commutators, and then finish it off by noting that conjugation by an element is a homomorphism.

For the first part, well, you have already done it.

The second part is even easier. It's straight forward to check that conjugation by an element gives a homomorphism. That's, $(xy)^z=x^zy^z$.

The result is now proved, because the product of commutators must get mapped to products of commutators. That's $[G,G]^z=[G,G]$ for any $z$. Note that I used your result that the inverse of a commutator is a commutator. Indeed, that means elements of the commutator are all products of commutators.

Answer 3

$x[a_1,b_1]\cdot...\cdot[a_n,b_n] x^{-1}= (x[a_1,b_1]x^{-1})\cdot ...\cdot (x[a_n,b_n]x^{-1})= \\ [xa_1x^{-1},xb_1x^{-1}]\cdot...\cdot [xa_nx^{-1},xb_nx^{-1}]\in [G,G]$

Answer 4

The elegant way to prove this result is as follows:

Lemma. Consider a group $G$ and two normal subgroups $K, H \trianglelefteq G$. Then the commutator subgroup $[K, H] \trianglelefteq G$ between $K$ and $H$ is also normal.

Proof. Let us abbreviate $F\colon=[K, H]$. Agreeing to denote the subgroup generated by an arbitrary subset $X \subseteq G$ by $\langle X \rangle$, let us recall the fact that given an arbitrary group morphism $f \colon G \to G'$ we have the relation: $$f[\langle X \rangle]=\langle f[X] \rangle \tag{gen}$$ for any subset $X \subseteq G$. Also note that $f([x, y])=[f(x), f(y)]$ for any $x, y \in G$.

I will use the notation ${}^tx\colon=txt^{-1}$ for left conjugates (the natural choice for notation, as left conjugation induces a left action of the group $G$ on the support set abusively referred to also as $G$). In order to prove that $F$ is normal, it suffices to show that ${}^tF \subseteq F$ for any $t \in G$. Introducing the subset $M\colon=\{[x, y]\}_{\substack{x \in K\\y \in H}}$, we have by definition that $F=\langle M \rangle$. It follows from the relation (gen) applied to the inner automorphism given by left conjugation with $t$ that: ${}^tF={}^t\langle M \rangle=\langle {}^tM \rangle \subseteq \langle M \rangle=F,$ the last of the inclusions being justified by the fact that ${}^tM \subseteq M$.

Indeed, for any $u \in M$ there exist by definition $x \in K$ and $y \in H$ such that $u=[x, y]$ and thus ${}^tu={}^t[x, y]=\left[{}^tx, {}^ty\right] \in M$, bearing in mind the normality of $K$ and $H$ (which ensures the fact that the conjugates ${}^tx \in K$ and ${}^ty \in H$ remain within the respective subgroups). $\Box$

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The more general setting in which this particular result can be framed is that normal subsets of groups generate normal groups, where by a normal subset $X \subseteq G$ I mean one such that ${}^tX=X$ for any $t \in G$. This is equivalent to claiming the validity of the relation ${}^GX\colon=\left\{{}^tx\right\}_{\substack{t \in G\\x \in X}} \subseteq X$.

Answer 5

You're missing a decisive step.

Lemma. Let $S$ be a subset of $G$ and $H$ be the subgroup generated by $S$. Then $H$ is normal in $G$ if and only if, for every $x\in S$ and $g\in G$, $gxg^{-1}\in H$.

Proof. One direction is obvious. For the other one, one has to use the fact that conjugation is an automorphism of the group (endomorphism suffices).

Denote by ${}^gx=gxg^{-1}$ and assume that ${}^gx\in H$, for all $x\in S$ and $g\in G$. Then consider the set $S^{-1}$ of inverses of members of $S$. Then it's easy to show that, for every $y\in S^{-1}$ and every $g\in G$, we have ${}^gy\in S^{-1}$. Since the subgroup $H$ consists of products of members of $S'=S\cup S^{-1}$ we just need to prove that, whenever $x_1,x_2,\dots,x_n$ are members of $S'$, then also ${}^g(x_1x_2\dotsm x_n)\in H$. This is an easy induction, after proving the $n=2$ case. □

In the case of $S$ being the set of commutators, you can observe that $[x,y]^{-1}=[y,x]$, so we just need to see that, for every $g,x,y\in G$, ${}^g[x,y]$ is a product of commutators. And now $$ {}^g[x,y]=[g,x][x,gy] $$ suffices.