교환자부분군

교환자부분군(commutator subgroup)에 대한 문제 하나를 풀어보았습니다.

문제

Let $G$ be a group and let $G’=\langle aba^{-1}b^{-1}\rangle$; that is, $G’$ is the subgroup of all finite products of elements in $G$ of the form $aba^{-1}b^{-1}$. The subgroup $G’$ is called the commutator subgroup of $G$.¹

(a) Show that $G’$ is a normal subgroup of $G$.

(b) Let $N$ be a normal subgroup of $G$. Prove that $G/N$ is abelian if and only if $N$ contains the commutator subgroup of $G$.

풀이

(a) Show that $G’$ is a normal subgroup of $G$.

We show that $gG’g^{-1} \subset G’$ for any $g\in G$. Let $x\in G’$, then $gxg^{-1}=x(x^{-1}gxg^{-1})$, and $x^{-1}gxg^{-1}$ is an element of $G’$ by the definition of the commutator subgroup. Thus $gxg^{-1}\in G’$ and so $gG’g^{-1}\subset G’$. Therefore, $G’$ is a normal subgroup of $G$.

(b) Let $N$ be a normal subgroup of $G$. Prove that $G/N$ is abelian if and only if $N$ contains the commutator subgroup of $G$.

$$\begin{align} G/N\text{ is abelian}&\Longleftrightarrow (aN)(bN)=(bN)(aN)\text{ for all }a,b\in G\\ &\Longleftrightarrow aba^{-1}b^{-1}\in N \text{ for all }a,b\in G\\ &\Longleftrightarrow G' \subset N \end{align}$$

각주

1. Thomas W. Judson, Robert A. Beezer. (2016). Abstract Algebra: Theory and Applications, p. 124. ↩