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Quotient groups

We now introduce a new construction for groups, which will enable us to prove that the kernels of homomorphisms are precisely normal subgroups.


\begin{thm}
Let $G$\ be a group, and let $N\unlhd G$.
On the set $G/N=\{ aN\st a...
...N)(bN)=(ab)N.
\end{displaymath}With this operation, $G/N$\ is a group.
\end{thm}


\begin{proof}
First we have to prove that the above multiplication is {\em well ...
...(a^{-1}a)N=(a^{-1}N)(aN).
\end{displaymath}This completes the proof.
\end{proof}


\begin{ex}
% latex2html id marker 3530Consider the quaternion group $G$\ and t...
...se
\ref{ex71}. (Note that $Q_8/N$\ is abelian, although $Q_8$\ is not.)
\end{ex}


\begin{exc}
Describe $S_n/A_n$.
\end{exc}

We return to the kernels again.


\begin{thm}
Let $G$\ be a group, and let $N\unlhd G$.
The mapping $f\st G\longrightarrow G/N$\ defined by
$f(x)=xN$\ is a homomorphism and $\ker(f)=N$.
\end{thm}


\begin{proof}
That $f$\ is a homomorphism follows from
\begin{displaymath}
f(xy)...
...x)=eN=N$.
In particular $x=xe\in N$. Therefore $\ker(f)\subseteq N$.
\end{proof}



Edmund F Robertson

11 September 2006