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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Problem Set 2}
\centerline{January 28, 2008}
\vskip 10pt
\noindent
{\it Section 8.2}
\vskip 10pt
\noindent
22. 
We prove the result first for finite abelian $p$-groups
for an arbitary prime $p$.  
Let $G$ and $H$ be two finite abelian $p$-groups such that for
all $m\ge 1$, the numbers of elements of order $m$
in $G$ and $H$ are the same.  The $m$'s involved
in this case are powers of $p$, of course. 
Since $G$ and $H$
are finite, they have the same number of
elements of the same maximal order $p^M$ for some $M$.
then writing
\[
N_i(G) = |\{g \in G : |g| = p^i\}|,
\]
(the number of elements of order $p^i$), we have
\[
|G| = \sum_{i=0}^M N_i(G) 
    = \sum_{i=0}^M N_i(H)= |H|.
\]

We will argue \emph{by induction} on the order of 
the groups that $G$ and $H$ must be isomorphic 
under the given hypothesis.  If
$|G| = |H| = 1$, then there is nothing to prove
since $G = \{0_G\}$ and $\{0_H\} = H$ are both
trivial groups hence are isomorphic. 

Now assume the result has been 
proved for $p$-groups of all orders $< p^k$ and consider
$G,H$ of order $p^k$ such that for all $m \le p^k$,  
the numbers of elements of order $m$
in $G$ and $H$ are the same. 
By hypothesis, the maximal
orders are the same, say $p^M$ ($M \le k$), 
and there are the same numbers of elements 
in $G$ and $H$ of order $p^M$.  Pick one such element 
$a \in G$ and one such element
$b \in H$.  By Lemma 8.6 (used
in the proof of the Structure Theorem), we have
$G = \langle a\rangle \oplus K$ and
$H = \langle b\rangle \oplus L$ 
for some subgroups $K,L$.
Now, $K,L$ are $p$-groups of order $p^{k-M}$.

Moreover, we claim that they must also satisfy
the hypothesis -- $K$ and $L$ must contain
the same number of elements of order $p^i$ for
each $p^i$.  We will prove this by a separate
induction argument.  The base case is $i = 0$
and the statement is true in that case since
there is only one element of order 1 in each group
(the $0$ elements).   Suppose we know that 
\[
N_j(K) = N_j(L), \quad \text{ for all } 0 \le j \le i-1.
\]
Consider the numbers of elements of order $p^i$ in
$K,L,G$, and $H$.  By the direct sum decompositions
for $G$, any element of order $p^i$ in $G$
must look like $b + c$ where $b \in \langle a\rangle$
and $c \in K$ and ${\rm lcm}(|b|,|c|) = p^i$ (that
is, at least one of $|b|,|c|$ is $p^i$ and the other
is $\le p^i$).  We count how many elements this gives,
writing $C = \langle a\rangle$:
\[
N_i(G) = (N_0(C) + \cdots + N_i(C))N_i(K) + (N_0(K) + \cdots + N_{i-1}(K))N_i(C).
\]
Similarly from the direct sum decomposition  $H = \langle b\rangle \oplus L = D\oplus L$ 
we get
\[
N_i(H) = (N_0(D) + \cdots + N_i(D))N_i(L) + (N_0(L) + \cdots + N_{i-1}(L))N_i(C).
\]
Now $C$ and $D$ are isomorphic since they are cyclic groups of the same
order.  This means $N_j(C) = N_j(D)$ for all $j$.  The induction
hypothesis shows that the terms in the second sets of parentheses
are equal.  Moreover $N_i(G) = N_i(H)$ by
hypothesis.  Therefore $N_i(K) = N_i(L)$ and this finishes the 
inner induction argument.     
This shows that $K$ and $L$ satisfy the same hypothesis as $G$ and $H$.

Hence by the induction hypothesis (for the outer induction), $K\simeq L$.
Since $C \simeq D$ as above, this shows that $G \simeq H$. 

Now, consider the general case, where $G,H$ are finite abelian groups,
but not necessarily $p$-groups.  The same reasoning used for $p$-groups
(but applied to elements of all possible orders) shows that $|G| = |H|$. 
 The hypothesis implies in particular
that for any prime $p$, the number of elements of order $p^i$ for 
each $i$ is the same in both $G$ and $H$.  Letting $i$ range over all
integers $i\ge 0$, these elements are exactly the 
elements of the subgroups $G(p)$ and $H(p)$.  The above proof shows
that $G(p) \simeq H(p)$.  This is true for all $p$ dividing the order
of $G,H$.  Hence by the Structure Theorem $G\simeq H$.  
\vskip 10pt
\noindent
{\it Section 8.3}
\vskip 10pt
\noindent
2. (a) Since $|S_4| = 24 = 2^3\cdot 3$, Sylow $2$-subgroups are 
subgroups of order $2^3 = 8$.   We can get one such subgroup
by taking the cyclic subgroup generated by the $4$-cycle $(1234)$
and including additional suitable elements of order $2$ to 
make subgroups isomorphic to $D_4$:
\begin{eqnarray*}
H_1 &=& \{(), (1234), (13)(24), (1432), (24), (13), (12)(34), (14)(23)\} 
\end{eqnarray*}
Then, other subgroups of the same order can be generated by conjugating
this one.  For instance conjugating by $(12)$ we get
\begin{eqnarray*}
H_2 &=& \{(), (1342), (23)(14), (1243), (14), (23), (12)(34), (24)(13)\}.
\end{eqnarray*}
Conjugating by $(23)$ we get
\begin{eqnarray*}
H_3 &=& \{(), (1324), (12)(34), (1423), (34), (12), (13)(24), (34)(12)\}.
\end{eqnarray*}

(b) The Sylow $3$-subgroups have order 3; they are cyclic, generated
by the 3-cycles:
\begin{eqnarray*}
K_1 &=& \{(), (123), (132)\}\\
K_2 &=& \{(), (124), (142)\}\\
K_3 &=& \{(), (134), (143)\}\\
K_4 &=& \{(), (234), (243)\}.
\end{eqnarray*}
\vskip 10pt
\noindent
3.  Since $|A_4| = 12 = 2^2 \cdot 3$, the Sylow $2$-subgroups 
have order $4$.  There is exactly one such subgroup:
\[
H = \{(),(12)(34),(13)(24),(14)(23)\}.
\]
The Sylow $3$-subgroups have order 3; they are cyclic, generated
by the 3-cycles (this is the same as in $S_4$):
\begin{eqnarray*}
K_1 &=& \{(), (123), (132)\}\\
K_2 &=& \{(), (124), (142)\}\\
K_3 &=& \{(), (134), (143)\}\\
K_4 &=& \{(), (234), (243)\}.
\end{eqnarray*}
{\it Comment:}  Note by Sylow 3 that the number of $2$-subgroups
must be an odd number dividing 12.  The possible numbers are 
1 and 3, but there are only 3 elements of order a power of 2 in $A_4$.  
\vskip 10pt
\noindent
4. The elementary divisors are $2,2^2,2^2,3,3,5$.   There is exactly
one Sylow $2$-subgroup, isomorphic to $Z_2\oplus Z_4 \oplus \Z_4$.
There is one Sylow $3$-subgroup, isomorphic to $\Z_3 \oplus \Z_3$.
There is one Sylow $5$-subgroup, isomorphic to $\Z_5$.
\vskip 10pt
\noindent
5.  (a) By Sylow theorem 3, the number of $3$-subgroups is congruent to 
$1$ mod $3$ and divides $72$.  The number could be 1 or 4. 
(b) Similarly, the number is congruent to $1$ mod $5$ and divides
$60$.  Hence the possibilities are $1$ or $6$. 
\vskip 10pt
\noindent
6. (b)  $143 = 11 \cdot 13$.  By Sylow 3, the number of Sylow $11$-subgroups
is $1$.  Similary, the number of $13$-subgroups is $1$.  By Sylow theorem 2, 
both of these subgroups are normal.  Hence $G \simeq H \times K$ is 
cyclic of order $143$.  
\vskip 10pt
\noindent
{\it Section 8.4}
\vskip 10pt
\noindent
1 (a)  Use the permutation form of $D_4$ from 8.3/2a above:
\[
D_4 \simeq \{(), (1234), (13)(24), (1432), (24), (13), (12)(34), (14)(23)\} \subset S_4
\]
to make computing conjugates easier(!) The conjugacy classes are
\begin{eqnarray*}
C_1 &=& C_{()} = \{()\} \quad \text{ in the center }\\
C_2 &=& C_{(1234)} = \{(1234),(1432)\}\\
C_3 &=& C_{(13)(24)} = \{(13)(24)\} \quad \text{ also in the center } \\ 
C_4 &=& C_{(12)(34)} = \{(12)(34),(14)(23)\}\\
C_5 &=& C_{(13)} = \{(13),(24)\}.
\end{eqnarray*} 
There is no real ``trick'' to this -- just compute!
\vskip 10pt
(c) The conjugacy classes in 
\[
A_4 = \{(),(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}
\]
are
\begin{eqnarray*}
C_1 &=& \{()\}\\
C_2 &=& \{(12)(34),(13)(24),(14)(23)\}\\
C_3 &=& \{(123),(243),(134),(142)\}\\
C_4 &=& \{(132),(234),(143),(124)\}.
\end{eqnarray*}
(Note in particular that the two elements of order 3 in each Sylow
3-subgroup end up in different conjugacy classes in $A_4$.  This
is true because the permutations that conjugate $(123)$ to $(132)$
in $S_4$, like $(23)$, are {\it odd} -- they are not in $A_4$.  Hence
the conjugacy classes in $S_4$ split up into smaller subsets 
in $A_4$.)
\end{document}
