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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.

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

First we have to prove that the above multiplication is {\em well ...
\end{displaymath}This completes the proof.

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

Describe $S_n/A_n$.

We return to the kernels again.

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 $\ker(f)=N$.

That $f$\ is a homomorphism follows from
In particular $x=xe\in N$. Therefore $\ker(f)\subseteq N$.

Edmund F Robertson

11 September 2006