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\begin{document}
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{\Large {\bf Homework 10}} \hfill MATH 301/601 \\
{Solutions to Graded Problems\\[-2mm]
\line(1,0){470}
\medskip
\begin{Exercise}[Section 9.4 \#22]
Let \( G \) be a group of order 20.
If \( G \) has subgroups \( H \) and \( K \) of orders 4 and 5, respectively, such that \( hk = kh \) for all \( h \in H \) and \( k \in K \), prove that \( G \) is the internal direct product of \( H \) and \( K \).
\end{Exercise}
\begin{Solution}
Since we are given that every element of \( H \) commutes with every element of \( K \), we only need to check that \( H \cap K = \{e \} \) and \( G = HK \).
Let us first show that the intersection is trivial.
If \( a \in H \cap K \), then by Lagrange's Theorem, the order of \( a \) must divide the order of \( H \) and also the order of \( K \).
As the orders of \( H \) and \( K \) are relatively prime, we must have that \( |a| = 1 \), and hence \( a = e \); in particular, \( H \cap K = \{e\} \).
Now, the set \( HK = \{ hk : h \in H, k \in K \} \) has at most \( |H|\cdot |K| \) elements; we claim it has exactly that many elements.
Suppose \( h, h' \in H \) and \( k, k' \in K \) such that \( hk = h'k' \).
Rearranging, this tells us that \( (h')^{-1}h = k' k^{-1} \).
The product on the left-hand side is in \( H \) and the product on the right-hand side is in \( K \).
But \( H \cap K =\{e\} \), so we must have that \( (h')^{-1}h = k'k^{-1} = e \), implying that \( h' = h \) and \( k' = k \).
This establishes that \( |HK| = |H|\cdot |K| \).
Therefore, as \( |H| \cdot |K| = |G| \), we have that \( |G| = |HK| \), implying \( G = HK \).
We have now shown that \( G \) is the internal direct product of \( H \) and \( K \).
\end{Solution}
\setcounter{Exercise}4
\medskip
\begin{ExChallenge}
Let \( N \) be a group, and let \( H \) be a subgroup of \( \rm{Aut}(N) \), the automorphism group of \( N \).
The \emph{(external) semidirect product} of \( N \) and \( H \) is the group \( N \rtimes H \) whose underlying set is \( N \times H \) and whose group operation is defined by \( (a, \varphi)(b, \psi) = (a\varphi(b), \varphi \circ \psi) \).
\begin{enumerate}[(a)]
\item Prove that \( N\rtimes H \) is a group (yes, I said it was a group in the definition, but that needs a proof).
\item Let \( G \) be a group, and let \( N \) and \( H \) be subgroups of \( G \) such that \
\begin{enumerate}[(i)]
\item \( N \cap H = \{e\} \),
\item \( G = NH = \{nh : n \in N, h \in H \} \), and
\item \( h n h^{-1} \in N \) for all \( n \in N \) and all \( h \in H \).
\end{enumerate}
(Note that condition (ii) and (iii) together imply that \( N \) is a normal subgroup of \( G \), see the Week 12 notes.)
Condition (iii) says that each element of \( H \) induces an automorphism of \( N \) via conjugation, that is, for \( h \in H \) we can define \( \varphi_h \in \rm{Aut}(N) \) by \( \varphi_h(n) = hnh^{-1} \) for all \( n \in N \).
Identifying \( h \) with \( \varphi_h \), we can view \( H \) as a subgroup of \( \rm{Aut}(N) \).
Under these hypotheses and under this identification of \( H \) with a subgropup of \( \rm{Aut}(N) \), prove that \( G \cong N \rtimes H \).
Here, we say \( G \) is the \emph{(internal) semidirect product} of \( N \) and \( H \).
\end{enumerate}
\end{ExChallenge}
\begin{Solution}
\begin{enumerate}[(a)]
\item Associativity follows from the associativity of the group operations of \( N \) and \( H \); I'll leave the details to you.
There is an identity element, namely \( (e_N, e_H) \).
It is left to check inverses.
If \( (a, \varphi) \in N \rtimes H \), then observe that \( (a, \varphi)(\varphi^{-1}(a^{-1}), \varphi^{-1}) = (e_N, e_H) \), so \( (a,\varphi)^{-1} = (\varphi^{-1}(a^{-1}), \varphi^{-1}) \).
\item
Define \( \Phi \co N \rtimes H \to G \) by \( \Phi(n,h) = nh \).
Condition (i) tells us that \( \Phi \) is injective, and condition (ii) tells us that \( \Phi \) is surjective.
To finish, let \( (n, h), (n',h') \in N\rtimes H \).
Then
\begin{align*}
\Phi((n,h)(n',h')) &= \Phi(nhn'h^{-1}, hh') \\
&= nhn'h^{-1}hh' \\
&= nhn'h' \\
&= \Phi(n,h)\Phi(n',h').
\end{align*}
All together, we have that \( \Phi \) is an isomorphism.
\end{enumerate}
\end{Solution}
\end{document}