next up previous contents
Next: Normal subgroups Up: MT2002 Algebra Previous: Homomorphisms of groups   Contents

Isomorphisms

Let us compare the group ${\mathbb{Z}}_3$ under addition modulo $3$ with the cyclic subgroup $H$ of $S_3$ generated by $(1\ 2\ 3)$:

\begin{displaymath}
\begin{array}{c\vert ccc}
{\mathbb{Z}}_3 & 0&1&2\\ \hline
0&...
...1\ 3\ 2)&\id\\
(1\ 3\ 2)&(1\ 3\ 2)&\id&(1\ 2\ 3)
\end{array}.
\end{displaymath}

We see that the two tables differ only in the names of the symbols, and not in their positions. More formally, there is mapping $f\st {\mathbb{Z}}_3\longrightarrow H$ (namely $f(0)=\id$, $f(1)=(1\ 2\ 3)$, $f(2)=(1\ 3\ 2)$) which is a bijection and satisfies $f(i+j)=f(i)\circ f(j)$.


\begin{de}
Let $G$\ and $H$\ be two groups.
A mapping $f\st G\longrightarrow H$\...
...s an
isomorphism $f\st G\longrightarrow H$; this is denoted $G\cong H$.
\end{de}


\begin{thm}
Let $G$, $H$\ and $K$\ be three groups. Then:
\begin{itemize}
\item[...
... (iii)]
$G\cong H\ \&\ H\cong K\Longrightarrow G\cong K$.
\end{itemize}\end{thm}


\begin{proof}
(i) The identity mapping on $G$\ is an isomorphism.
(ii) If $f\st ...
...isms, then so is their composition
$g\circ f\st G\longrightarrow K$.
\end{proof}

It makes sense to regard isomorphic groups as identical. The main general task of group theory can be formulated as: classify all non-isomorphic groups. In general this is impossible, and one has to settle for various partial results in this direction. Probably the easiest such is the following:


\begin{thm}
Let $G$\ be a cyclic group.
If $G$\ is infinite then $G\cong{\mathbb{Z}}$;
if $G$\ is finite of order $n$\ then $G\cong {\mathbb{Z}}_n$.
\end{thm}


\begin{proof}
% latex2html id marker 3411Let $a$\ be a generator for $G$.
Assu...
...ightarrow G$by $f(i)=a^i$. As above, this mapping is an isomorphism.
\end{proof}

In order to prove that two groups $g$ and $H$ are not isomorphic, one needs to demonstrate that there is no isomorphism from $G$ onto $H$. Usually, in practice, this is much easier than it sounds in general, and is accomplished by finding some property that holds in one group, but not in the other.


\begin{ex}
The groups ${\mathbb{Z}}_4$\ and ${\mathbb{Z}}_6$\ are not isomorphic because they have
different orders.
\end{ex}


\begin{ex}
${\mathbb{Z}}_6\not\cong S_3$\ because ${\mathbb{Z}}_6$\ is abelian and
$S_3$\ is not (although they both have order $6$).
\end{ex}


\begin{ex}
The dihedral group $D_{12}$\ is not isomorphic to $S_4$because $D_{12...
...rt D_{12}\vert=\vert S_4\vert=24$\ and that they are both non-abelian.)
\end{ex}


\begin{exc}
Consider the group $Q_8=\{ 1,-1,i,-i,j,-j,k,-k\}$,
where the multipl...
...}.
Prove that $Q_8$\ is not isomorphic to $D_4$\ and ${\mathbb{Z}}_8$.
\end{exc}

Finally, we prove the so called Cayley's theorem, which suggests a prominent role of the symmetric groups among all groups.


\begin{thm}[Cayley]
Every group $G$\ is isomorphic to a subgroup of the symmetric group $S(G)$.
\end{thm}


\begin{proof}
% latex2html id marker 3445For each $a\in G$\ define a mapping $...
... it also becomes onto, and
we conclude that $G\cong\im(f)\leq S(G)$.
\end{proof}


\begin{re}
The permutations $\tau_a$\ can be read off easily from the
Cayley tab...
... an arbitrary fashion),
and thus represent $G$\ as a subgroup of $S_n$.
\end{re}


\begin{ex}
In the quaternion group $Q_8$, we have, for example,
\begin{displayma...
...1&2&3&4&5&6&7&8\\
3&4&2&1&7&8&6&5
\end{array}\right).
\end{displaymath}\end{ex}

The practical use of Cayley's Theorem is limited: it is not very likely that one can obtain much useful information about groups of order, say, eight, by considering subgroups of the group $S_8$ of order $40320$.


next up previous contents
Next: Normal subgroups Up: MT2002 Algebra Previous: Homomorphisms of groups   Contents

Edmund F Robertson

11 September 2006