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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Discussion 3 -- Alternating groups; Cayley's Theorem}
\centerline{November 26, 2007}
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\noindent
{\it Background}
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Before the break, we saw that the symmetric group $S_n$ contains
a normal subgroup $A_n$ whose elements are all of the ``even''
permutations, the permutations $\sigma$ whose corresponding 
$n\times n$ permutation
matrices $\rho(\sigma)$ have determinant 1:
$$A_n = \{\sigma \in S_n : \det(\rho(\sigma)) = 1\}.$$
$A_n$ is called the {\it alternating group} on $n$ letters.
Today we will study the symmetric and alternating groups
and develop some more of their properties.
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\noindent
{\it Discussion Questions}
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\noindent
I. 
\item{A)}  List all of the permutations in $A_3$ and $A_4$ using disjoint
cycle decomposition notation.
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\item{B)}  Determine the group tables for $A_3$ and $A_4$.  ({\it Comment:} The
table for $A_4$ will be
slightly tedious.  You can ``parallelize'' the computation (split it up 
among your group members), though.   
This is an important group we have not seen before, 
so it is worth the effort!) 
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\item{C)}  One fairly common {\it misconception} about Lagrange's Theorem
is to think that if $k$ divides $|G|$, then there must be a subgroup $H \subset G$
with $|H| = k$.  But is that true for all of the divisors of $|A_4|$?  Are
there subgroups of $A_4$ of all the possible orders allowed by Lagrange's Theorem?
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\item{D)}  You should have seen that $A_3$ and $A_4$
have half the number of elements of $S_3$ and $S_4$ respectively.
Prove in general that $|A_n| = {1\over 2} n!$.   
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\noindent
II.  One of the reasons that the symmetric groups are important 
examples is a famous result called {\it Cayley's Theorem}.  In 
intuitive terms, this result says that if we understand the 
symmetric groups and their subgroups, then we understand 
{\it every finite group} $G$, because {\it every finite
group $G$ is isomorphic to a subgroup of $S_n$ for some $n$}.
In this problem, you will look at some examples and work through 
one proof of this theorem.

Let $G$ be a finite group of order $n$ and list the elements as
$$G = \{g_1,g_2,\ldots,g_n\}\leqno(1)$$
in some order (the exact order is immaterial -- any order will work as
long as you maintain it consistently).  As we know from a question 
on the last discussion, each row and column of the group table
for $G$ contains each element of $G$ exactly once.  If we look at
the columns, this means that the {\it left multiplication mappings}
$$\eqalign{L_g : G &\to G\cr
                 x &\mapsto gx\cr}$$
are permutations of $G$ (one-to-one, onto mappings from $G$ to itself).
We can identify each one with a permutation of $\{1,2,\ldots,n\}$
via the listing of the group elements from (1) above.  
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\item{A)} Let $G$ be a cyclic group of order $4$, say with $a$ as 
generator.  Write down the permutation of $G$ defined by $L_g$
for each element of $G$, both in disjoint cycle notation, and
as a permutation matrix.   
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\item{B)}  Generalize your results from part A to cyclic groups
of order $n$ for any integer $n \ge 2$.
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\item{C)} Repeat part A for $G = S_3$.  (Be careful! What size will
the permutation matrices be?)
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\noindent
Now we want to develop a proof of:

\proclaim Cayley's Theorem.  Every finite group $G$ is isomorphic
to a subgroup of $S_n$ for some $n$.  \par

\noindent
The $n$ here will actually be $n = |G|$.  We introduce the following
notation:  Let $\widetilde{G}$ be the set of all the left multiplication
mappings $L_g : G \to G$ by elements of $G$:
$$\widetilde{G} = \{L_g : g \in G\},$$
viewed as permutations in $S_n$.  
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\item{D)}  Show that $\widetilde{G}$ is a group under the operation
of composition of functions.  Hence it can be viewed as a subgroup of $S_n$
by identifying each $L_g$ mapping with its corresponding permutation
in $S_n$.
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\item{E)}  Show that the mapping $\varphi : G \to \widetilde{G}$ defined
by $\varphi(g) = L_g$ is an 
{\it isomorphism of groups}.  This completes the proof of Cayley's Theorem.
\end
