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% This is for the title
{\Large {\bf Homework 11}} \hfill MATH 301/601 \\
{Solutions to Graded Problems\\[-2mm]
\line(1,0){470}

\medskip







\setcounter{Exercise}3
\medskip
\begin{Exercise}
Let \( G \) be a cyclic group, let \( a \) be a generator of \( G \), and let \( \varphi, \psi \co G \to H \) be homomorphisms. Prove that if \( \varphi(a) = \psi(a) \), then \( \varphi = \psi \). 
(This says that a homomorphism defined on a cyclic group is completely determined by its action on a generator of the group.)
\end{Exercise}


\begin{Solution}
In order to show that the functions \( \varphi \) and \( \psi \) are equal, we need to show that \( \varphi(g) = \psi(g) \) for each \( g \in G \). 
So, let \( g \in G \).
As \( a \) is a generator of \( G \), there exists \( n \in \bz \) such that \( g = a^n \). 
We have previously shown that isomorphisms respect exponentiation, so we have:
\begin{align*}
\varphi(g) 	&= \varphi(a^n) \\
		&= \varphi(a)^n \\
		&= \psi(a)^n \\
		&= \psi(a^n) \\
		&= \psi(g).
\end{align*}
As \( g \) was an arbitrary element of \( G \), we have established that \( \varphi = \psi \). 
\end{Solution}

\setcounter{Exercise}5
\medskip
\begin{ExChallenge}
The subgroup of a group \( G \) generated by the set \( \{ xyx^{-1}y^{-1} : x,y \in G \} \) is called the \emph{commutator subgroup of \( G \)} and is denoted \( G' \) (or \( [G,G] \)). 
\begin{enumerate}[(a)]
\item Prove that \( G' \) is normal in \( G \).
\item Prove that \( G/G' \) is abelian.
\item Let \( N \) be a normal subgroup of \( G \). Prove that \( G/N \) is abelian if and only if \( G' \subset N \). 
\end{enumerate}
(The group \( G/G' \) is called the \emph{abelianization} of \( G \).)
\end{ExChallenge}

\begin{Solution}
\begin{enumerate}[(a)]

\item 
The product \( xyx^{-1}y^{-1} \) is called a \emph{commutator} and is usually denoted by \( [x,y] \). 
It is readily verified that \( g[x,y]g^{-1} = [gxg^{-1}, gyg^{-1}] \).
This says that the generating set of \( G' \) is invariant under conjugation, and hence \( G' \) is invariant under conjugation, i.e., \( gG'g^{-1} = G \); in other words, \( G' \) is normal.

\item 
There is a bit of redundancy in parts (b) and (c), so let us prove the following: If \( N \) is a normal subgroup of \( G \) containing \( G' \), then \( G/N \) is abelian.
To see this, let \( a,b \in G \).
We need to show that \( (ab)N = (ba)N \). 
From one of our lemmas, we know that \( (ab)N = (ba)N \) if and only if \( (ba)^{-1}(ab) \in N \). 
Here, \( (ba)^{-1}(ab) = a^{-1}b^{-1}ab = [a^{-1}, b^{-1}] \in G' \), and as \( G' < N \), we have that \( (ab)N = (ba)N \).
We have established that \( G/N \) is abelian. 

\item
We have already established the reverse direction, so assume that \( G/N \) is abelian.
We want to show that \( G' < N \). 
By definition, \( G' \) is the smallest subgroup that contains every commutator, so to show that \( G' < N \), we need to show that \( N \) contains every commutator. 
Let \( x, y \in G \). 
As \( G/N \) is abelian, we have that \( (x^{-1}y^{-1})N = (y^{-1}x^{-1})N \).
This implies that \( (y^{-1}x^{-1})^{-1}(x^{-1}y^{-1}) \in N \), or equivalently, \( [x,y] \in N \). 
As \( x \) and \( y \) were arbitrary elements of \( G \), we have shown that \( N \) contains every commutator of elements in \( G \), and hence it must contain \( G' \). 
\end{enumerate}
\end{Solution}


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