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\noindent{\Large{\textsc{Modern Algebra \hfill Assignment 6.1  \\ Spring 2019 \hfill Due February 27}}}

\large

\vspace{1cm}

\begin{exercise}
Let $A$ be an (additive) abelian group, $B_1, B_2, \ldots, B_n < A$.  Define
\[
f : B_1 \times B_2 \times \cdots \times B_n \to A
\]
by $f(b_1, b_2, \ldots, b_n) = b_1 + b_2 + \cdots + b_n$.

\begin{itemize}

\item[\bf a.]  Prove that $f$ is a homomorphism.  Use this to conclude that
\[
B_1 + B_2 + \cdots + B_n = \{ b_1 + b_2 + \cdots + b_n \, | \, b_i \in B_i \}
\]
is a subgroup of $A$.

\item[\bf b.]  Prove that $\ker f = \{ 0 \}$ if and only if $(B_1 + B_2 + \cdots + \widehat{B_i} + \cdots + B_n) \cap B_i = \{ 0 \}$ for all $1 \le i \le n$.  Here the caret indicates the omission of the summand $B_i$.  

\item[\bf c.]  Prove that $\ker f = \{ 0 \}$ if and only if $B_i \cap (B_{i+1}+\cdots + B_n) = \{ 0 \}$ for all $1 \le i \le n-1$.

\end{itemize}

\noindent  When $\ker f = \{ 0 \}$, $f$ provides an isomorphism between the {\em external} product $B_1 \times B_2 \times \cdots \times B_n$ and the {\em internal} sum $B_1 + B_2 + \cdots + B_n$.  In this case we call the latter the {\em (internal) direct sum} and denote it by $B_1 \oplus B_2 \oplus \cdots \oplus B_n$.  

\end{exercise}

\begin{exercise}
With $A$ and $B_i$ as above, prove that
\[
B_1 \oplus B_2 \oplus B_3 = B_1 \oplus (B_2 \oplus B_3) = (B_1 \oplus B_2) \oplus B_3,
\]
i.e.\ the direct sum is associative.  Yes, there really is something to prove here. 
\end{exercise}

\begin{exercise}
Let $f:G \to H$ be a group homomorphism.

\begin{itemize}

\item[\bf a.]  Prove that for any $a \in G$, the order of $f(a)$ divides the order of $a$.  

\item[\bf b.]  Prove that if $f$ is a monomorphism, then for any $a \in G$, $a$ and $f(a)$ have the same order.  

\item[\bf c.]  Use part {\bf b} (and an earlier result) to provide an alternate proof of the fact that for any $a, b \in G$, $a$ and $bab^{-1}$ have the same order.

\end{itemize}

\end{exercise}

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