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{\Large {\bf Homework 5}} \hfill MATH 301/601 \\
{Solutions to Graded Problems\\[-2mm]
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\medskip


\setcounter{Exercise}2


\bigskip
\begin{Exercise}
Suppose \( H \) is a nonempty finite subset of a group \( G \) and that \( H \) is closed under multiplication (that is, \( ab \in H \) for all \( a, b \in H \)). 
Prove that \( H \) is a subgroup of \( G \). 
\end{Exercise}

\begin{Solution}
As \( H \) is closed under multiplication, it is enough to show that \( H \) contains the identity of \( G \) and that \( H \) is closed under inversion.
We know \( H \) is nonempty, so let \( h \in H \). 
Using the fact that \( H \) is finite, we can find \( j,k \in \bz \) such that \( j < k \) and \( h^j = h^k \). 
It follows that \( h^{k-j} = e \). 
Now, as \( H \) is closed under multiplication and \( k-j \in \bn \), we have that \( e = h^{k-j} \in H \).
Also, observe that \( h^{k-j-1} = h^{-1} \), and as \( k-j-1 \in \bn \cup \{0\} \), we have that \( h^{-1} \in H \).
As \( h \) was an arbitrary element of \( H \), we see that \( H \) is closed under taking inverses. 
Therefore, \( H \) is a subgroup of \( G \). 
\end{Solution}


\setcounter{Exercise}5


\medskip
\begin{Exercise}
Suppose \( G \) is a nontrivial group in which the only two subgroups of \( G \) are itself and the trivial subgroup.
\begin{enumerate}[(a)]
\item Prove that \( G \) is cyclic.
\item Using part (a), prove that \( G \) is a finite group of prime order.
\end{enumerate}
\end{Exercise}

\begin{Solution}
(a) As \( G \) is nontrivial, there exists \( g \in G \ssm \{e\} \). 
Since \( g \) is not the identity, the cyclic subgroup generated by \( g \) is nontrivial, and hence, as \( G \) has only two subgroups, the cyclic subgroup generated by \( g \) must be all of \( G \).
Therefore, \( G \) is cyclic. 
In fact, we have shown that every element of \( G \ssm \{e\} \) is a generator of \( G \). 

(b) We have already established that \( G \) is cyclic, so we may choose a generator of \( G \), call it \( a \). 
If \( a^2 = e \), then \( |G| = 2 \), and hence \( G \) is a finite group of prime order. 
We can therefore assume that \( a^2 \neq e \), and hence, the cyclic subgroup generated by \( a^2 \) is equal to \( G \).
Therefore, there exists \( k \in \bz \) such that \( (a^2)^k = a \), which implies that \( a^{2k-1} = e \).
Hence, \( G \) has finite order.

Now, suppose that \( n \in \bn \ssm \{1\} \) divides \( |G| \).
We want to show that \( n = |G| \).
Let \( q \in \bz \) such that \( |G| = nq \).
As \( n > 1 \), we must have that \( q < |G| \); hence,  \( a^q \neq e \) and \( |a^q| = |G| \).
But, \( (a^q)^n = (a^{qn}) = a^{|G|} = e \), which implies that \( n \geq |G| \).
Therefore, as \( n \mid |G| \), we must have that \( n = |G| \).
We have shown that the only divisors of \( |G| \) are 1 and itself, and so \( |G| \) is prime. 
\end{Solution}

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