# A Condition that a Commutator Group is a Normal Subgroup ## Problem 3

Let $H$ be a normal subgroup of a group $G$.
Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$.

Here $[H, G]$ is a subgroup of $G$ generated by commutators $[h,k]:=hkh^{-1}k^{-1}$.

In particular, the commutator subgroup $[G, G]$ is a normal subgroup of $G$ Add to solve later

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## Proof.

First, we show that $N=[H, G]$ is a subgroup of $H$.
A generator of $N$ is of the form either $hgh^{-1} g^{-1}$ or $ghg^{-1} h^{-1}$, where $h \in H$ and $g \in G$. Since $H$ is normal in $G$, we see that these are elements in $H$. Thus $N <H$.

Next, we show that $N$ is normal in $G$.

Let $x=hgh^{-1} g^{-1}$ be a generator element of $N$.

Then for any $a \in G$, we have
$ax a^{-1} =ahgh^{-1} g^{-1}a^{-1}=(aha^{-1})(aga^{-1})(ah^{-1}a^{-1})(a g^{-1}a^{-1}) \in [H, G].$ Similarly, if $x$ is a generator of the form $ghg^{-1} h^{-1}$, then we see that $ghg^{-1} h^{-1} \in [H, G]$ by the same argument.
Thus the conjugate of a generator of $N$ by $a \in G$ stays in $N$.

Now any element $x \in N$ is of the form $x=x_1 x_2 \cdots x_n$, where $x_i$ are generators of $N$.

For any $a \in H$, we have
$axa^{-1}=a x_1 x_2 \cdots x_n a^{-1}= (ax_1 a^{-1})(a x_2 a^{-1}) \cdots (a x_n a^{-1}) \in [H, G].$ Thus $N \triangleleft G$.

In particular, we apply the result to $H=G$. Then we see that the commutator subgroup $[G, G]$ is a normal subgroup of $G$.

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### 2 Responses

1. 10/01/2016

[…] Another problem about a commutator group is A condition that a commutator group is a normal subgroup […]

2. 10/18/2016

[…] The commutator subgroup $D(G)=[G,G]$ is a normal subgroup of $G$. For a proof, see: A condition that a commutator group is a normal subgroup. […]

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