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\noindent{\Large{\textsc{Modern Algebra \hfill Assignment 13.1  \\ Spring 2023 \hfill Due April 19}}}

\large

\vspace{1cm}


\begin{exercise}
Let $A$ be an additive abelian group and let $A_1, A_2, \ldots, A_n \le A$. Let
\[
A' = A_1 + A_2 + \cdots + A_n  =\{a_1 + a_2 + \cdots + a_n \, | \, a_i \in A_i \text{ for all } i \}.
\]
Define $f: A_1 \times A_2 \times \cdots \times A_n \to A$ by 
\[
f(a_1, a_2, \ldots, a_n) = a_1 + a_2 + \cdots + a_n.
\]
\begin{itemize}

\item[\bf a.]  Prove that $f$ is a homomorphism and that $\operatorname{im} f = A'$.  Conclude that $A'$ is a subgroup of $A$.

\item[\bf b.]  Prove that the following conditions are equivalent.

\begin{itemize}

\item[\bf i.]  $f: A_1 \times A_2 \times \cdots \times A_n \to A'$ is an isomorphism.

\item[\bf ii.]  If $a_i, b_i \in A_i$ for all $i$ and
\[
a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n,
\]
then $a_i = b_i$ for all $i$.

\item[\bf iii.]  For all $i$, we have
\[
A_i \cap \sum_{j \ne i} A_j = \{ 0 \}.
\]

\end{itemize}

\end{itemize}

\noindent When the (equivalent) conditions of part {\bf b} hold, we say that the sum of the $A_i$ is {\em direct}, and we write
\[
A_1 + A_2 + \cdots + A_n = A_1 \oplus A_2 \oplus \cdots \oplus A_n.
\]

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\begin{exercise}
Lang, II.4.4
\end{exercise}


\begin{exercise}
Lang, II.4.28
\end{exercise}


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