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\noindent{\Large{\textsc{Modern Algebra \hfill Assignment 11.1  \\ Spring 2019 \hfill Due April 17}}}

\large

\vspace{1cm}


\begin{exercise}
Let $A$ and $B$ be abelian groups, and let $f:A \to B$ be a homomorphism.  Suppose there is a homomorphism $g : B \to A$ so that $f \circ g = 1_B$.  Prove that $A = \ker f \oplus \operatorname{im} g$.  [{\em Suggestion:}  Given $a \in A$, write $a = (a - g(f(a))) + g(f(a))$.]
\end{exercise}

\begin{exercise}  
Let $A$ be an additive finite cyclic group of order $n$.  For each $m \in \Z$ with $\gcd(m,n)=1$ define $\sigma_m : A \to A$ by $\sigma_m(a) = ma$.  

\begin{itemize}

\item[\bf a.]  Prove that $\sigma_m \in \operatorname{Aut}(A)$.  [{\em Suggestion:}  Use the fact that $\gcd(m,n)=1$ to show that $\ker \sigma_m = \{0\}$.  Since $A$ is finite, this implies $\sigma_m$ is onto (why?).]

\item[\bf b.]  Prove that $\sigma_m = \sigma_{\ell}$ if and only if $m \equiv \ell \pmod{n}$.  [{\em Suggestion:}  For one implication, apply the two automorphisms to a generator of $A$.]

\item[\bf c.]  Use parts {\bf a} and {\bf b} to construct a monomorphism $(\Z/n\Z)^{\times} \to \operatorname{Aut}(A)$.  [{\em Remark:}  This is {\em not} just a consequence of the first isomorphism theorem, since $(\Z/n\Z)^{\times}$ is not a quotient of $\Z$, but a subset of a quotient.]

\item[\bf d.]  Show that the map of part {\bf d} is an isomorphism.  [{\em Suggestion:}  If $A$ is generated by $x$, and $\sigma \in \operatorname{Aut}(A)$, then $\sigma(x) = mx$ for some integer $m$.  Use the fact that $|\sigma(x)| = |x|$ to show that $\gcd(m,n) = 1$.]

\end{itemize}

\end{exercise}

\noindent {\bf Remark.}  A somewhat more natural, but more advanced, approach to Exercise 2 that {\em does} take advantage of the first isomorphism theorem replaces $\operatorname{Aut}(A)$ with $\operatorname{End}(A)$, the {\em ring} of arbitrary homomorphisms $A \to A$.  The map $m \mapsto \sigma_m$ yields a surjective ring homomorphism $\Z \to \operatorname{End}(A)$, and the ring-theoretic first isomorphism theorem then implies $\Z/n\Z \cong \operatorname{End}(A)$.  By restricting this to the {\em unit groups} on both sides, we get the isomorphism of {\bf 2d} above ``for free.''



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