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\noindent{\Large{\textsc{Modern Algebra \hfill Assignment 12.1  \\ Spring 2023 \hfill Due April 12}}}

\large

\vspace{1cm}


\begin{exercise}
Let $C$ be an (additive) cyclic group of order $n$, generated by $a$, and suppose $n = de$ for some $d, e \in \N$.  Prove that
\[
C[d] = \left\langle e a \right\rangle.
\]
\end{exercise}

\begin{exercise}  
Let $G$ be a group and let $\mathcal{H}$ be a {\em directed} set of subgroups of $G$: for all $H, K \in \mathcal{H}$ there exists a $J \in \mathcal{H}$ so that $H \cup K \le J$.  Prove that
\[
L = \bigcup_{H \in \mathcal{H}} H
\]
is a subgroup of $G$.
\end{exercise}

\begin{exercise}
An (additive) abelian group $A$ is said to be a {\em torsion} group if every element of $A$ has finite order.  Given a prime $p$ define
\[
A(p) = \bigcup_{e=1}^{\infty} A[p^e],
\]
which consists of all the elements of $A$ whose order is some power of $p$.

\begin{itemize}

\item[\bf a.]  Use the preceding exercise to show that $A(p) \le A$.

\item[\bf b.]  Show that
\[
A = \sum_{p} A(p),
\] 
where the (internal) sum runs over all primes $p$.\footnote{The {\em sum} of a (possibly infinite) collection $\mathcal{B}$ of subgroups of $A$ is defined to be the subgroup of $A$ generated by $S=\bigcup_{B \in \mathcal{B}} B$.  It consists of all {\em finite} sums of elements in $S$.} [{\em Suggestion.}  If $0 \ne a \in A$ and $n = |a|$, then $a \in A[n]$.  Express $A[n]$ as the internal direct sum of prime-power-torsion subgroups $A[p^e]$, and use the fact that $A[p^e] \le A(p)$.]

\item[\bf c.]  Show that the internal sum $\sum_p A(p)$ is direct.  [{\em Suggestion.}  Suppose $a_i \in A(p_i)$ for some primes $p_1, p_2, \ldots, p_r$ and that $a_1 + \cdots + a_r = 0$.  Then for each $i$, $a_i \in A[p_i^{e_i}]$ for some $e_i \ge 1$.  But the sum $A[p_1^{e_1}] + \cdots + A[p_r^{e_r}]$ is direct (why?), so that $a_i = 0$ for all $i$.]

\end{itemize}




\end{exercise}


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