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Cosets and Lagrange's Theorem

Every subgroup of a group induces an important decomposition of .

$\begin{de} Let $G$\ be a group, let $H$\ be a subgroup of $G$, and let $a\in G$\... ...a$\ is the set \begin{displaymath}Ha=\{ha\st h\in H\}. \end{displaymath}\end{de}$

$\begin{ex} Let us find all the (left and right) cosets of the cyclic subgroup $H... ... 3)\}& R_6=H(1\ 3\ 2)=\{(1\ 3\ 2),(1\ 3)\}. \end{array}\end{displaymath}\end{ex}$

$\begin{exc} Write down the cosets of the subgroup $\{0,3,6,9\}$\ in ${\mathbb{Z}}_{12}$. \end{exc}$

$\begin{exc} What are the cosets of the trivial subgroup $\{e\}$\ in $G$? What are the cosets of $G$\ in $G$? \end{exc}$

$\begin{exc} Prove that the cosets of $A_n$\ in $S_n$\ are $A_n$\ and $O_n$. \end{exc}$

We see that the left and the right coset determined by the same element need not be equal. We may also notice some interesting regularities: all the cosets have equal sizes; some of them are equal (e.g. ), and the others are disjoint (e.g. $L_3\cap L_6=\emptyset$ ). These are not accidents.

$\begin{thm} Let $G$\ be a group, and let $H$\ be a subgroup of $G$. \begin{itemi... ... equal size.) \end{itemize}Analogous statements hold for right cosets. \end{thm}$

$\begin{proof} (i) $H=eH$. \par (ii) $a=ae\in aH$, since $e\in H$. \par (iii) Cle... ...ghtarrow ax=ay\Rightarrow x=y, \end{displaymath}and $f$\ is one-one. \end{proof}$

$\begin{thm}[Lagrange] Let $G$\ be a group of finite order, and let $H$\ be a subgroup of $G$. Then the order of $H$\ divides the order of $G$. \end{thm}$

$\begin{proof} % latex2html id marker 2170Let $C_1,\ldots,C_k$\ be the distinct... ...t H\vert}_k=k\vert H\vert, \end{displaymath}and the theorem follows. \end{proof}$

$\begin{thm} Let $G$\ be a finite group, and let $a\in G$\ be arbitrary. Then the order of $a$\ divides the order of $G$. \end{thm}$

$\begin{proof} The order of $a$\ is equal to the order of the cyclic subgroup of $G$generated by $a$. \end{proof}$

We give one interesting application.

$\begin{thm} Every group of prime order is cyclic. \end{thm}$

$\begin{proof} Let $G$\ be a group of prime order $p$, and let $a\in G$\ be any n... ...$p$, so that $n=p$. It follows that $G$\ is cyclic generated by $a$. \end{proof}$

Edmund F Robertson

11 September 2006