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\title{Conjugacy classes of the Alternating Group $A_n$}
\author{GCS: 17-xi-03}
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This note is to correct some insanity in Friday's lecture.
In what follows, $n$ is a positive integer and at least 2.

First we recall a general fact which we proved via
working on a problem sheet:

\noindent{\bf Lemma } Suppose that $G$ is a group and
that $H$ and $K$ are subgroups of $G$, then 
$|K:H \cap K| \leq |G:H|$.

Suppose that $x \in A_n \leq S_n$. Let the
conjugacy class of $x$ in $S_n$ be $C$ and 
the conjugacy class of $x$ in $A_n$ be $D$.

\noindent{\bf Theorem }\ $C = D$ if and only 
$x$ commutes with an odd element of $S_n$. If 
this is not the case, then $2|D| = |C|.$

\noindent{\bf Proof }\ Apply the lemma with
$S_n = G$, $A_n = H$ and $C_{S_n}(x) = K.$ 
Note that $C_{A_n}(x) = A_n \cap C_{S_n}(x)$
so $|C_{S_n}(x):C_{A_n}(x)| \leq |S_n:A_n| = 2.$

Now if this index is 1 if and only if  $C_{S_n}(x)
= C_{A_n}(x)$ which happens if and only if $x$
commutes with no odd element of $S_n$. 
In these circumstances $|S_n:A_n| \cdot |A_n:C_{A_n}(x)|
= |S_n:C_{A_n}(x)| = |S_n:C_{S_n}(x)|$ and so
$2|D| = |C|.$

On the other hand if $|C_{S_n}(x): C_{A_n}(x)|=2$, then a factor
of two can be cancelled from 
$|S_n:A_n| \cdot |A_n: C_{A_n}(x)|
= |S_n:C_{A_n}(x)| = |S_n:C_{S_n}(x)|\cdot |C_{S_n}(x):C_{A_n}|$
to yield that $|C| = |D|$, but $D \subseteq C$ so $C = D$. Conversely
if $C = D$, then $|S_n:C_{S_n}(x)| = |A_n:C_{A_n}(x)|$.
Now $|S_n:A_n| \cdot |A_n: C_{A_n}(x)|
= |S_n:C_{A_n}(x)| = |S_n:C_{S_n}(x)|\cdot |C_{S_n}(x):C_{A_n}|$
so $|S_n:A_n| = |C_{S_n}(x):C_{A_n}| =2$.

The proof is complete.

\noindent{\bf Example }\ In $S_3$ the elements 
$(1,2,3)$ and $(1,3,2)$ are conjugate.
In $A_3 = \langle (1,2,3) \rangle$ they are not (the group $A_3$
is abelian so different elements are never conjugate). The conjugacy class
of $(1,2,3)$ is $S_3$ is $\{ (1,2,3), (1,3,2) \}$ whereas 
is $A_3$ it is $\{ (1,2,3) \}$. Notice that the centralizer of
$(1,2,3)$ in $S_3$ is $\langle (1,2,3) \rangle$ so $(1,2,3)$
commutes with no odd elements of $S_3$.

\noindent{\bf Observations}\ In the lecture I speculated that when a conjugacy class of $S_n$ 
falls into two conjugacy classes in $A_n$, then the elements of 
one class might necessarily be the inverses of 
the elements of the other. Note that
inverse elements do have the same cycle shape since you can invert
an element in standard form by reversing each of its cycles.
This speculation turns out to be false.
In $A_5$ there are two conjugacy classes of elements with shape
which is a 5-cycle. However $g = (2,5)(3,4) \in A_5$
and $g^{-1}(1,2,3,4,5)g = (1,5,4,3,2) = (1,2,3,4,5)^{-1}.$ 
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