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Cosets and Lagrange's Theorem

Every subgroup of a group $G$ induces an important decomposition of $G$.


\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. $L_3=L_5$), 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