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Normal subgroups

We have seen that to every homomorphism $f\st G\longrightarrow H$ we can associate two distinguished subgroups: $\ker(f)\leq G$ and $\im(f)\leq H$. It is natural to ask a converse question: given a subgroup of a group $G$, is this subgroup the kernel or the image of some homomorphism?

It is easy that every subgroup is the image of some homomorphism. Indeed, if $K\leq G$, define $f\st K\longrightarrow G$ by $f(x)=x$. Then it is clear that $\im(f)=K$.

The situation for kernels is different. In order to describe it, we introduce a special class of subgroups.


\begin{thm}
The following conditions are equivalent for a subgroup $N$\ of a gro...
...ry $a\in G$\ we have $aNa^{-1}=\{a^{-1}na\st n\in N\}=N$.
\end{itemize}\end{thm}


\begin{proof}
% latex2html id marker 3473(i)$\Rightarrow$(ii)
Let $a\in G$. By...
...usion. Hence $aN=Na$,
and so every left coset is also a right coset.
\end{proof}


\begin{de}
A subgroup $N$\ of a group $G$\ is {\em normal} if it satisfies any (...
...quivalent conditions of the above theorem; this is denoted $N\unlhd G$.
\end{de}


\begin{ex}
$\{e\}\unlhd G$; $G\unlhd G$, as left and right cosets are obviously equal.
\end{ex}


\begin{ex}
Every subgroup of an abelian group is normal,
since $aN=Na$\ is obviously satisfied.
\end{ex}


\begin{ex}
% latex2html id marker 3494The cyclic subgroup of $S_3$\ generated ...
...ause the left and right cosets do not coincide;
see Example \ref{ex51}.
\end{ex}


\begin{ex}
% latex2html id marker 3497If $N$\ is a subgroup of $G$\ with exact...
...s N$\ as well.
In particular $A_n\unlhd S_n$; see Exercise \ref{exc52}.
\end{ex}


\begin{ex}
% latex2html id marker 3500Consider the quaternion group $Q_8$; see...
...$Q_8$.
Hence we have $aN=Na$\ for every $a\in Q_8$, and $N$\ is normal.
\end{ex}

We now return to our investigation of the kernels.


\begin{thm}
If $f\st G\longrightarrow H$\ is a homomorphism then
$\ker(f)\unlhd G$.
\end{thm}


\begin{proof}
Let $a\in G$\ and $n\in\ker(f)$. From $f(n)=e_H$\ it follows that
...
...(a)=
f(a)^{-1}f(a)=e_H.
\end{displaymath}Hence $a^{-1}na\in\ker(f)$.
\end{proof}



Edmund F Robertson

11 September 2006