Let $G$ a group and $X$ a subset of $G$. Let $X^G=\langle X^a : a \in G\rangle$, where $X^a=aXa^{-1}$. Show (using the conjugation classes theory) that $G$ is a minimal normal subroup of $G$ containing $X$.
Using conjugation classes
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0Do you mean to show that $X^G$ is a minimal normal subgroup of $G$ containing $X$? – 2012-10-16
1 Answers
You have three things to prove:
- $X\subseteq X^G$;
- $X^G\unlhd G$; and
- if $X\subseteq N\unlhd G$, then $X^G\le N$.
I’ll leave (1) to you. For (2), suppose that $y\in X^G$, $g\in G$, and $y^g\notin X^G$. By definition $X^G=\bigcap\left\{H\le G:\bigcup_{a\in G}X^a\subseteq H\right\}\;,$
so $G$ has a subgroup $H$ such that $\bigcup_{a\in G}X^a\subseteq H$ and $y^g\notin H$. Then $y=\left(y^g\right)^{g^{-1}}\notin H^{g^{-1}}$, and
$H^{g^{-1}}\supseteq\left(\bigcup_{a\in G}X^a\right)^{g^{-1}}=\bigcup_{a\in G}X^{ag^{-1}}=\bigcup_{a\in G}X^a\;,$ so $X^G\subseteq H^{g^{-1}}\subseteq G\setminus\{y\}$, and therefore $y\notin X^G$, contradicting the choice of $y$. Thus, $y^g\in X^G$ for all $y\in X^G$ and $g\in G$, and therefore $X^G\unlhd G$.
For (3), suppose that $X\subseteq N$, where $N\unlhd G$. Then $X^a\subseteq N^a$ for every $a\in G$, so ... ?
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0@Jarbas: Let $\mathscr{H}$ be the collection of subgroups of $G$ that contain the set $\bigcup_{a\in G}X^a$; then $X^G=\bigcap\mathscr{H}$. If $y^g$ were in $H$ for every $H\in\mathscr{H}$, then $y^g$ would be an element of $X^G$. But $y^g\notin X^G$, so $y^g$ cannot belong to every $H\in\mathscr{H}$. That is, there must be at least one $H\in\mathscr{H}$ such that $y^g\notin H$. – 2012-10-16