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{\Large {\bf Homework 4}} \hfill MATH 301/601 \\
{Solutions to Graded Problems\\[-2mm]
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\bigskip
\begin{Exercise}[\#32 in Section 3.5]
Show that if \( G \) is a finite group of even order, then there is an \( a \in G \) such that \( a \) is not the identity and \( a^2 = e \). 
\end{Exercise}

\begin{Solution}
Let \( E = \{ g \in G : g^{-1} \neq g \} \).
Pairing each element of \( E \) with its inverse, we see that \( E \) has an even number of elements.  
As \( |G| = |E| + |G\ssm E| \) and both \( |G| \) and \( |E| \) are even,  \( G \ssm E \) has an even number of elements. 
Note if \( g \in G\ssm E \), then \( g = g^{-1} \), and hence \( g^2 = e \).
Therefore, to finish, we need to show that \( G \ssm E \) contains a non-identity element.
We know that \( e \in G \ssm E \), so \( G\ssm E \) has at least one element.
But, \( G \ssm E \) is even and hence has at least two elements.
Therefore, there exists \( a \in G \ssm E \) such that \( a \neq e \).  
As noted above, this means that \( a^2 = e \), and \( a \) is our desired element of \( G \). 
\end{Solution}


\setcounter{Exercise}2


\medskip
\begin{ExChallenge}
Let \( G \) be a finite group.
Prove that there exists \( N \in \bn \) such that \( g^N = e \) for each \( g \in G \). 
\end{ExChallenge}

\begin{Solution}
Fix \( g \in G \). 
Consider the subset \( \{ g^k : k \in \bn \} \subset G \). 
As \( G \) is finite, the above subset has only finitely many elements, and hence, there exists \( i,j \in \bn \) such that \( g^j = g^i \) and \( j > i \). 
It follows that \( g^jg^{-i} = e \), and setting \( N_g = j-i  \), we have that \( N_g \in \bn \) and  \( g^{N_g} = e \). 

Now, let \( N = \prod_{g\in G} N_g \), and, for \( g \in G \), let \( N'_g = N/N_g = \prod_{h \in G \ssm \{g\}} N_h  \in \bn  \). 
We then have that, for any \( g \in G \), 
\[
g^N = g^{N_gN'_g} = \left( g^{N_g} \right)^{N'_g} = e ^{N'_g} = e,
\]
and hence \( N \) is as desired. 
\end{Solution}



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