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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Problem Set 1}
\centerline{January 21, 2008}
\vskip 10pt
\noindent
{\it Section 8.2}
\vskip 10pt
\noindent
1.  Let $G$ be an abelian group. 
By definition, $G(p) = \{a \in G : p^\ell \cdot a = 0 \text{\ for some\ }
\ell \ge 0\}$.  Let $a,b \in G(p)$.  Then $p^\ell\cdot a = 0$
and $p^m\cdot b = 0$ for some $\ell,m\ge 0$.  Let $n = \max\{\ell,m\}$.
Then $p^n\cdot a = p^n\cdot b = 0$ so
\[ 
p^n\cdot(a - b) = p^n\cdot a - p^n \cdot b = 0 - 0 = 0. 
\]
This shows that $a - b \in G(p)$.  Hence $G(p)$ is a subgroup of $G$.
\vskip 10pt
\noindent
2.  Let $G$ be an abelian group and let $pG = \{p\cdot x : x \in G\}$.
Let $a = p\cdot x$ and $b = p\cdot y$ be elements of $pG$, where $x,y \in G$.
Then $a - b = p\cdot x - p\cdot y = p\cdot (x - y)$ and $x - y \in G$ since
$G$ is a group under $+$.  Hence $a - b \in pG$, so $pG$ is a subgroup.
\vskip 10pt
\noindent
3. (d) We have $72 = 2^3 \cdot 3^2$.  As in the example on p. 256 in
Hungerford or the examples done in class last Friday, the structure
theorem implies every abelian group of order 72 has elementary divisors
$2^3, 3^2$ or $2^3, 3, 3$, or $2^2, 2, 3^2$, or $2^2, 2, 3, 3$, or
$2,2,2,3^2$, or $2,2,2,3,3$.  The group is isomorphic to one of 
\begin{eqnarray*}
\Z_8 \oplus \Z_9, &\quad & \Z_8 \oplus \Z_3 \oplus \Z_3, \\
\Z_4 \oplus \Z_2 \oplus \Z_9, &\quad & \Z_4 \oplus \Z_2 \oplus \Z_3 \oplus \Z_3,\\
\Z_2 \oplus \Z_2 \oplus \Z_2 \oplus \Z_9, &\text{\ or\ }& \Z_2 \oplus \Z_2  \oplus \Z_2 \oplus \Z_3 \oplus \Z_3.
\end{eqnarray*}

\vskip 10pt
\noindent
(h)  Similarly, $1160 = 2^3\cdot 5 \cdot 29$, so every abelian group
of order 1160 is isomorphic to one of 
\[
\Z_8 \oplus \Z_5 \oplus \Z_{29},\  \Z_4 \oplus \Z_2 \oplus \Z_5 \oplus \Z_{29},
\text{\ or\ } \Z_2 \oplus Z_2 \oplus \Z_2 \oplus \Z_5 \oplus \Z_{29}.
\]
\vskip 10pt
\noindent
5 (b)  We can use the fact that if $(m,n) = 1$, then 
$\Z_{mn} \simeq \Z_m \oplus \Z_n$ (Lemma 8.8).  Applying
this, 
\[
\Z_6 \oplus \Z_{12} \oplus \Z_{18} \simeq (\Z_2 \oplus \Z_3) \oplus
(\Z_4 \oplus \Z_3) \oplus (\Z_2 \oplus \Z_9).
\]
Since the elementary divisors are unique, this shows the elementary
divisors of  $\Z_6 \oplus \Z_{12} \oplus \Z_{18}$
are $2,2,4,3,3,9$ (reordering to put powers in increasing order).
\vskip 10pt
\noindent
{\it Comment:}  Note that this also gives the invariant factors of 
this group, which is isomorphic to $\Z_6 \oplus \Z_6 \oplus \Z_{36}$.
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\noindent
8. By definition,
\[
G(2) = \left\{\left[\frac{m}{n}\right] \in \Q/\Z : 2^\ell \cdot \left[\frac{m}{n}\right] = [0] \text{\ for some\ }
\ell \ge 0\right\}.
\]
This is equivalent to saying that $2^\ell\frac{m}{n} $ is an integer.  If
$\frac{m}{n}$ is written in lowest terms, then $n = 2^\ell$ for some $\ell\ge 0$,
and hence $G(2)$ is the set of cosets of rational numbers
of the form $\frac{m}{2^{\ell}}$ in lowest terms.   Similarly, $G(p)$ is the 
set of cosets of rational numbers of the form $\frac{m}{p^\ell}$
in lowest terms.
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\noindent
9. (a) Saying $G$ is a $p$-group means by definition that every 
element of $G$ has order $p^\ell$ for some $\ell \ge 0$.  If
$G$ is also a finite group, then there is some element $m \in G$
of \emph{maximal} order $p^{\ell_0}$ where $\ell_0 \ge 1$.  
Now consider the group $pG$ as in Exercise 2 above.  If 
$x$ has order $p^\ell$ with $\ell \ge 1$, then $p^{\ell-1} \cdot (p\cdot x) = 
p^\ell\cdot x = 0$, but $p^k \cdot (p\cdot x) = p^{k+1}\cdot x \ne 0$ 
if $k < \ell - 1$.  Hence the order of $p\cdot x$ is $\ell - 1$.  
This shows that $pG$ cannot contain any elements of order $\ell_0$.
Hence $pG \ne G$.  
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\noindent
(b)  Let $G = \Q/\Z$ as in Exercise 8.  $G(2)$ is the subgroup
consisting of cosets of rational numbers of the form 
$\frac{m}{2^\ell}$ for all $\ell \ge 0$.  Note that $G(2)$ is
not a finite group because $\ell$ can be arbitrarily large.
And moreover, if $x = \left[\frac{m}{2^\ell}\right] \in G(2)$
then $x = 2\cdot \left[\frac{m}{2^{\ell+1}}\right]$ and
$y = \left[\frac{m}{2^{\ell+1}}\right] \in G(2)$.  This shows 
that $G(2) \subseteq 2\cdot G(2)$.  The other inclusion is clear,
so in fact $G(2) = 2\cdot G(2)$.   
\vskip 10pt
\noindent
11.  By the structure theorem (Theorem 8.7), $G$
is isomorphic to a direct sum of subgroups, each of order
$q^e$ for some primes $q$ and exponents $e \ge 1$.  If 
$G$ is a $p$-group, then only powers of one prime can appear
among the elementary divisors, and 
\begin{equation}\label{pgp}
G \simeq \Z_{p^{e_1}} \oplus \cdots \oplus \Z_{p^{e_s}}
\end{equation}
for some $e_i \ge 1$.  What we must show is that if $pG = \{0\}$,
then all the $e_i = 1$.  Assume not.  That is, suppose $e_i \ge 2$
for some $i$.  Then let $a$ be a generator of $Z_{p^{e_i}}$, so 
$|a| = p^{e_i} > p$.  In the direct sum we have the element
$x = (0,\ldots,0,a,0,\ldots,0)$ ($a$ in the $i$th factor).  Then 
$p\cdot x = (0,\ldots,0,p\cdot a,0,\ldots,0) \ne (0,\ldots,0)$.
Hence the group $pG \ne \{0\}$.  By contraposition, if $pG = \{0\}$,
then all $e_i = 1$ in Eq.~(\ref{pgp}).
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\noindent
12.  Let $G$ be a finite abelian group and let $p\ |\ |G|$.  
By the structure theorem (Theorem 8.7), $G$ is isomorphic 
to a direct sum of cyclic groups of prime power orders.  
Each of those orders must divide $|G|$ and hence at least 
one must be a power of $p$ (since otherwise the order
of $G$ would be a product of powers of primes different from $p$
and that would contradict $p\ |\ |G|$).  Hence 
\[
G\simeq \Z_{p^k} \oplus H_2 \oplus \cdots \oplus H_s,
\]
for some $k \ge 1$ and some cyclic groups $H_2,\ldots,H_s$ of prime power orders.
Hence $G$ contains a subgroup isomorphic to $\Z_{p^k}$
for some $k \ge 1$.  Let $a$ be a generator of this subgroup, 
so $|a| = p^k$.  By the same sort of argument used in Exercise 9 (a) above,
we have $|p^{k-1}\cdot a| = p$ (also see part (4) of Theorem 7.8). 
Hence $G$ contains an element of order $p$.  
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\noindent
14.  We are given that
\begin{equation}\label{fac1}
|G| = p^t m \qquad  \text{where\ } (p,m) = 1.
\end{equation}
By Theorem 8.5, we also have
$G \simeq G(p) \oplus H$ where $H$ is the direct sum of the 
$G(q)$ for the primes $q \ne p$ dividing $|G|$, which implies
$|G| = |G(p)| \cdot |H|$.  By Lemma 8.6, it follows
that $|G(p)| = p^s$ for some $s$ and $|H|$ is a product of powers
of primes $q\ne p$. Hence
\begin{equation}\label{fac2}
|G| = p^s m' \qquad  \text{where\ } (p,m') = 1.
\end{equation}
 By unique factorization in $\Z$, comparing Eq.~(\ref{fac1}) 
and Eq.~(\ref{fac2}), we see that $t = s$, and $m = m'$.
Therefore, $|G(p)| = p^t$ which is what we wanted to show.
\vskip 10pt
\noindent
15.  If $|G| = p^t m$ with $(p,m) = 1$, then Exercise 14
shows that $G$ contains the subgroup $G(p)$ of order $p^t$.
In other words, the statement to be proved is true for $n = t$, the
\emph{largest} power of $p$ dividing $|G|$.  The statement is 
also clearly true with $n = 0$, since $\{0\}$ is a subgroup of $G$.

What remains to be
proved is that the same statement is true for all $n$ with
$1 \le n < t$ as well.   It suffices to show that $G(p)$
contains a subgroup of order $p^n$ since a subgroup of $G(p)$
is also a subgroup of $G$.  We know 
\[
G(p) \simeq \Z_{p^{e_1}} \oplus \cdots \oplus \Z_{p^{e_s}}
\]
for some $e_i \ge 1$ satisfying $e_1 + \cdots + e_s = t$.
The cyclic group $\Z_{p^e}$ contains
elements of orders $1,p,p^2,\ldots,p^e$ since if $a$ is a generator
with order $p^e$, then $p^{e-j}\cdot a$ has order $p^j$ whenever
$0 \le j \le e$.  
Now, if $1 \le n < t$, then we can always write $n = f_1 + \cdots + f_s$
where $0 \le f_i \le e_i$ for all $i$.  (In fact, we always can do this in 
many different ways when $s > 1$).  If $a_i$
is a generator of $\Z_{p^{e_i}}$, then that summand 
contains a subgroup $K_i = \langle p^{e_i - f_i}\cdot a_i\rangle$ 
of order $p^{f_i}$ by the observation above.
Hence $G(p)$ contains the subgroup $K = K_1 \oplus \cdots \oplus K_s$
which has order $p^{f_1 + \cdots + f_s} = p^n$.  

\vskip 10pt
\noindent
16.  We claim that this is equivalent to
saying that $n = p_1p_2 \cdots p_s$ for some \emph{distinct} primes
$p_1,\ldots,p_s$.  Such integers are called \emph{square-free}
numbers.  If $n$ has this form, then there is just one 
possibility for the elementary divisors of $G$, namely
the set $p_1,p_2,\ldots,p_s$. 
By Lemma 8.8 (applied repeatedly), we have
\[
G \simeq \Z_{p_1} \oplus \cdots \oplus \Z_{p_s} \simeq \Z_{p_1\cdots p_s} = \Z_n,
\]
and there is just one abelian group of order $n$ up to isomorphism.
Conversely, suppose there is just one abelian group of order $n$ 
up to isomorphism.  If any prime $p$ satisfies $p^2\ |\ |G|$, 
then there are at least two different possibilities for the elementary
divisors of $G$ and hence there are at least two nonisomorphic
abelian groups of order $n$.  Hence for each prime $p$ that divides
$|G|$, we must have $p^2\ \not|\ |G|$.  Hence $n$ is square-free. 
\end{document}
