Exercise 67.4

Question

The \emph{	extbf{order}} of an element $a$ of an abelian group $G$ is the smallest positive integer $m$ such that $ma = 0$, if such exists; otherwise, the order of $a$ is said to be infinite. The order of $a$ thus equals the order of the subgroup generated by $a$.
\begin{enumerate}[label=(\alph*)]
\item Show the elements of finite order in $G$ form a subgroup of $G$,
  called its \emph{	extbf{torsion subgroup}}.
\item Show that if $G$ is free abelian, it has no elements of finite order.
\item Show the additive group of rationals has no elements of finite order, but is not free abelian.
\end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$]
  It suffices to show the torsion elements $T(G)$ of $G$ is closed under sums and inverses. If $a,b \in T(G)$ with orders $m,n$ respectively, then $mn(a+b) = n(ma) + m(nb) = 0 \implies a+b \in T(G)$; likewise, $m(-a) = -(ma) = 0 \implies -a \in T(G)$.
\end{proof}
\begin{proof}[Proof of $(b)$]
  Any $0 
e a \in G$ is of the form $a = \sum k_\alpha a_\alpha$ where $\{a_\alpha\}$ is our basis of $G$ and $k_\alpha \in \mathbb{Z}$. If $ma = 0$, then $mk_\alpha a_\alpha = 0$ for all $\alpha$, which is a contradiction since each $a_\alpha$ also has infinite order.
\end{proof}
\begin{proof}[Proof of $(c)$]
  Suppose $\mathbb{Q} = \braket{a_\alpha}$, $\{a_\alpha\}$ a basis. Then for any $a_\alpha$, $a_\alpha/2$ must be of the form $a_\alpha/2 = \sum k_\alpha a_\alpha$ for $k_\alpha \in \mathbb{Z}$. But then, $a_\alpha = 2\sum k_\alpha a_\alpha$, and since $a_\alpha$ is a basis element, we have $a_\alpha = 2k_\alpha a_\alpha$. This implies $k_\alpha = 1/2$, which is a contradiction.
\end{proof}

Exercise 67.4

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Exercise 67.4

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