Why Is Every Subgroup Containing the Commutator Subgroup Normal?

A subgroup need not be normal in its group. But if it contains all commutators, conjugating one of its elements cannot take that element outside the subgroup. The reason is that the change caused by conjugation is itself a commutator.

The commutator subgroup

For elements $x,y$ of a group $G$, define the commutator

$$[x,y]=xyx^{-1}y^{-1}.$$

The commutator subgroup, also called the derived subgroup, is

$$[G,G]=\langle [x,y]:x,y\in G\rangle,$$

the subgroup generated by every commutator in $G$. It measures how far $G$ is from being abelian: the quotient $G/[G,G]$ is always abelian, and $G$ itself is abelian exactly when $[G,G]$ is the identity subgroup.

The claim is that if $H\leq G$ and $[G,G]\subseteq H$, then $H$ is normal in $G$. Normality means that $gHg^{-1}=H$ for every $g\in G$.

Proof by conjugation

Take any $g\in G$ and $h\in H$. By the definition of a commutator,

$$[g,h]=ghg^{-1}h^{-1}.$$

Rearranging gives

$$ghg^{-1}=[g,h]h.$$

The commutator $[g,h]$ lies in $[G,G]$, and $[G,G]\subseteq H$. Also, $h\in H$. Since $H$ is a subgroup, it is closed under products, so $[g,h]h\in H$. Therefore $ghg^{-1}\in H$ for all $g\in G$ and $h\in H$. This proves $gHg^{-1}\subseteq H$ for every $g$.

Applying the same inclusion with $g^{-1}$ in place of $g$ gives the reverse inclusion, so $gHg^{-1}=H$. Hence $H\trianglelefteq G$.

The quotient-group viewpoint

There is a shorter way to see the same fact. The quotient $G/[G,G]$ is abelian. If $H$ contains $[G,G]$, then $H/[G,G]$ is a subgroup of this quotient. Every subgroup of an abelian group is normal, so $H/[G,G]$ is normal in $G/[G,G]$. The subgroup correspondence theorem then implies that $H$ is normal in $G$.

This viewpoint also explains why the commutator subgroup appears: it is precisely what must be factored out to make the quotient abelian. Any subgroup containing it comes from a subgroup of that abelian quotient.

Worked example: subgroups of $S_3$ containing its commutator subgroup

The symmetric group $S_3$ is nonabelian, and its commutator subgroup is $A_3$, the subgroup of even permutations. One way to verify this is that the sign map $S_3\to\{1,-1\}$ has abelian image, so every commutator is even and $[S_3,S_3]\subseteq A_3$. In the other direction, a commutator of two transpositions is a nonidentity 3-cycle; its powers generate $A_3$. Thus $[S_3,S_3]=A_3$.

Now suppose $H\leq S_3$ and $A_3\subseteq H$. The quotient $S_3/A_3$ has order 2, so its only subgroups are the identity subgroup and the whole quotient. Subgroups $H$ containing $A_3$ correspond to those two quotient subgroups. Therefore $H=A_3$ or $H=S_3$, and both are normal in $S_3$.

For a direct check, conjugating an element of $A_3$ by any permutation still gives an even permutation, so it remains in $A_3$. The theorem guarantees this normality without requiring a separate conjugation check for each element.

Common mistakes

• Confusing containment with equality. The theorem does not say $H=[G,G]$. It says any subgroup between $[G,G]$ and $G$ is normal.
• Assuming the converse. A normal subgroup need not contain $[G,G]$. For example, the identity subgroup is normal in $S_3$, but it does not contain the nontrivial subgroup $A_3=[S_3,S_3]$.
• Using the wrong commutator convention. Some books define commutators with inverses in a different order. The proof still works, but write the conjugation identity using the convention in your text. With $[g,h]=ghg^{-1}h^{-1}$, the identity is $ghg^{-1}=[g,h]h$.
• Forgetting that $H$ must be a subgroup. Containing all commutators is not by itself enough for an arbitrary subset to be normal; the closure properties of a subgroup are used when multiplying $[g,h]$ by $h$.