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\noindent{\Large{\textsc{Modern Algebra \hfill Assignment 5.3  \\ Spring 2023 \hfill Due February 15}}}

\large

\vspace{1cm}

\begin{exercise}
Let $G$ be a group and let $S$ be a set.  Suppose that $f : G \to S$ is a bijection.  We can use $f$ to turn $S$ into a group in a natural way.  Given $x,y \in S$, we use $f^{-1}$ to pull them back to $G$, apply the binary operation in $G$, and then push the result forward to $S$.  That is, we define
\[
x \ast y = f(f^{-1}(x)\cdot f^{-1}(y)).
\]
Prove that $(S,\ast)$ is a group and that $f:(G,\cdot) \to (S,\ast)$ is an isomorphism. 
\end{exercise}

\begin{exercise}
Let $(G,\cdot)$ be a multiplicative group.  Given $a \in G$, recall that the left translation $\lambda_a :G \to G$ given by $\lambda_a(x) = ax$ is a bijection.  In general $\lambda_a$ is not an isomorphism.  However, if we let $\ast$ denote the binary operation on $G$ defined by taking $S=G$ and $f = \lambda_a$ in the preceding exercise, then $\lambda_a :(G,\cdot) \to (G,\ast)$ {\em is} an isomorphism.  Show that $\ast$ agrees with the operation introduced in exercise 1.2.3.
\end{exercise}

\noindent{\bf Remark.}  Exercise 2 shows that every group $G$ is {\em homogeneous}.  For any $a \in G$ we can use $\lambda_a$ to translate the original group structure on $G$ to another group structure on $G$ that uses $a$ as the identity, and $\lambda_a$ provides an isomorphism between the two.  So $G$ ``looks the same'' after left translation by any of its elements.  This means that the set theoretic nature of $G$ is irrelevant to any group structure imposed upon $G$.  Put another way, if you know that $G$ is a group, but you've forgetten its binary operation, there is no intrinsic way to recover it.  

\begin{exercise}
If $G$ is a multiplicative group and $H,K < G$, let
\[
HK = \{ab \, | a \in H, b \in K \}.
\]
This is the multiplicative analogue of the sum of two subgroups of an abelian group.

\begin{itemize}

\item[\bf a.] Give an example to show that if $G$ is nonabelian, then $HK$ is not necessarily a subgroup of $G$.

\item[\bf b.]  Suppose that $xHx^{-1} \subseteq H$ for all $x \in G$.  Show that $HK = KH$ and that $HK < G$.

\end{itemize}

\end{exercise}


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