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% This is for the title
{\Large {\bf Homework 9}} \hfill MATH 301/601 \\
{Solutions to Graded Problems\\[-2mm]
\line(1,0){470}

\medskip




\setcounter{Exercise}1
\medskip
\begin{Exercise}
Let \( G \) be a finite abelian group of order \( n \).
Suppose \( m \in \bn \) is relatively prime to \( n \). 
Prove that \( \varphi \co G \to G \) given by \( \varphi(g) = g^m \) is an automorphism of \( G \). 
(This says that every element of \( G \) has an \( m^{\text{th}} \)-root.)
\end{Exercise}

\begin{Solution}
We need to show that \( \varphi \) is bijective and that \( \varphi(ab) = \varphi(a)\varphi(b) \) for all \( a, b \in G \). 
Let us consider the latter condition first.  
If \( a, b \in G \), then 
\begin{align*}
\varphi(ab) 	&= (ab)^m \\
			&= (a^m)(b^m) \\
			&= \varphi(a)\varphi(b),
\end{align*}
where the second equality is using the fact that \( G \) is abelian.

Now, since \( G \) is finite and both the domain and codomain of \( \varphi \) are \( G \), we know that if \( \varphi \) is injective or  surjective, then it is bijective.
So, it is enough to show either injectivity or surjectivity; however, I will give an argument for each of three as reference.

First, let us consider injectivity.  
Let \( a,b \in G \).
If \( \varphi(a) = \varphi(b) \), then \( a^m = b^m \), implying that \( a^mb^{-m} = e \).  
Using the standard exponent laws and the fact that \( G \) is abelian, we can rewrite this equation as \( (ab^{-1})^m = e \). 
This equality, together with Lagrange's Theorem, tells us that \( |ab^{-1}| \) divides \( m \).
Lagrange's Theorem also tells us that \( |ab^{-1}| \) divides \( n \). 
Now, as \( n \) and \( m \) are relatively prime, we can conclude that \( |ab^{-1}| = e \), and hence \( ab^{-1} = e \).
Therefore, \( a = b \), and \( \varphi \) is injective.

Next, let us consider surjectivity. 
Let \( g \in G \). 
We must show that \( g \) is in the range of \( \varphi \). 
As \( m \) and \( n \) are relatively prime, there exists \( s,t \in \bz \) such that \( 1 = ms + nt \). 
Observe:
\begin{align*}
	g	&= g^1 \\
		&= g^{ms +nt} \\
		&= (g^s)^m (g^n)^t \\
		&= (g^s)^m \\
		&= \varphi(g^s),
\end{align*}
where the fourth equality uses the fact that \( g^n = e \), as \( n = |G| \). 
Hence, \( \varphi \) is surjective. 
\end{Solution}

\setcounter{Exercise}5

\medskip
\begin{ExChallenge}
Let \( \bq \) denote the group \( (\bq, +) \), and let \( \bq^\times \) denote the group \( (\bq\smallsetminus\{0\}, \cdot) \). 
\begin{enumerate}[(a)]
\item
Let \( \varphi \co \bq \to \bq \) be an isomorphism.
Prove that \( \varphi(x) = x\cdot \varphi(1) \) for all \( x \in \bq \). 
(This is saying that every automorphism of \( \bq \) is \( \bq \)-linear.)
\item
Use part (a) to prove that if \( \varphi \co \bq \to \bq \) is an isomorphism, then there exists \( q \in \bq \smallsetminus \{0\} \) such that \( \varphi(x) = qx \) for all \( x \in \bq \). 
\item
Use part (b) to prove that \( \mathrm{Aut}(\bq) \cong \bq^\times \). 
\end{enumerate}
\end{ExChallenge}

\begin{Solution}
\begin{enumerate}[(a)]
\item 
Let \( \frac ab \in \bq \). 
Recall that \( \bq \) is a group under addition. 
This means that \( \varphi(nq) = n \varphi(q) \) for any \( n \in \bz \). 
We can therefore write \( \varphi( a/b) = a \varphi(1/b) \). 
And also, \[ b \varphi(1/b) = b \varphi(1/b) = \varphi(b\cdot 1/b) = \varphi(1) ,\]
and hence \( \varphi(1/b) = \frac1b\varphi(1) \).
Substituting back into the equation above yields \( \varphi(a/b) = \frac ab \varphi(1) \), as desired. 

\item 
Let \( q = \varphi(1) \).
Part (1) immediately implies that \( \varphi(x) = qx \). 
Moreover, as \( \varphi \) is an automorphism, we must have that \( q \neq 0 \). 

\item 
Given \( q \in \bq^\times \), let \( \varphi_q \in \mathrm{Aut}(\bq) \) be defined by \( \varphi_q(x) = qx \). 
Define \( \Phi \co \bq^\times \to \mathrm{Aut}(\bq) \) by \( \Phi(q) = \varphi_q \).
We claim that \( \Phi \) is an isomorphism. 
First, we see it is injective: if \( \Phi(q) = \Phi(q') \), then \( q = \varphi_q(1) = \varphi_{q'}(1) = q' \). 
Now, surjectivity follows from part (b): if \( \varphi \in \mathrm{Aut}(\bq) \), then there exists \( q \in \bq^\times \) such that \( \varphi = \varphi_q \), and hence \( \varphi = \Phi(q) \). 
Finally, we need to check that \( \Phi(qq') = \Phi(q) \circ \Phi(q') \) for all \( q,q' \in \bq^\times \). 
Let \( q,q \in \bq^\times \), and let \( x \in \bq \).
We compute:
\begin{align*}
\Phi(q q')(x) &= \varphi_{qq'}(x) \\ 
&= (qq')x = q(q'x) \\ 
&= q\varphi_{q'}(x) \\ 
&= \varphi_q(\varphi_{q'}(x)) \\
&= (\varphi_q\circ\varphi_{q'})(x) \\
&= (\Phi(q) \circ \Phi(q'))(x),
\end{align*}
and hence \( \Phi(qq') = \Phi(q) \circ \Phi(q') \). 
\end{enumerate}
\end{Solution}

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