Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group

Problem 307

Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.

(a) Prove that $T(A)$ is a subgroup of $A$.

(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion elements.)

(b) Prove that the quotient group $G=A/T(A)$ is a torsion-free abelian group. That is, the only element of $G$ that has finite order is the identity element.

We write the group operation multiplicatively.
Let $x, y\in T(A)$. Then $x, y$ have finite order, hence there exists positive integers $m, n$ such that $x^m=e, y^n=e$, where $e$ is the identity element of $A$. Then we have
\begin{align*}
(xy)^{mn}&=x^{mn}y^{mn} \qquad \text{ (since $A$ is abelian)}\\
&=(x^m)^n(y^m)^n=e^me^n=e.
\end{align*}
Therefore the element $xy$ has also finite order, hence $xy \in T(A)$.

Also, we have
\begin{align*}
(x^{-1})^m=(x^m)^{-1}=e^{-1}=e.
\end{align*}
Hence the inverse $x^{-1}$ of $x$ has finite order, hence $x^{-1}\in T(A)$.

Therefore, the subset $T(A)$ is closed under group operation and inverse, hence $T(A)$ is a subgroup of $A$.

(b) $A/T(A)$ is a torsion-free abelian group

Since $A$ is an abelian group, the quotient $G=A/T(A)$ is also an abelian group.
For $a\in A$, let $\bar{a}=aT(A)$ be an element of $G=A/T(A)$. Suppose that $\bar{a}$ has finite order in $G$. We want to prove that $\bar{a}=\bar{e}$ the identity element of $G$.

Since $\bar{a}$ has finite order, there exists a positive integer $n$ such that
\[\bar{a}^n=\bar{e}.\]
This implies that
\[a^nT(A)=T(A)\]
and thus $a^n\in T(A)$.

Since each element of $T(A)$ has finite order by definition, there exists a positive integer $m$ such that $(a^n)^m=e$.
It follows from $a^{nm}=e$ that $a$ has finite order, and thus $a\in T(A)$.
Therefore we have
\[\bar{a}=aT(A)=T(A)=\bar{e}.\]

We have proved that any element of $G=A/T(A)$ that has finite order is the identity, hence $G$ is the torsion-free abelian subgroup of $G$.

Quotient Group of Abelian Group is Abelian
Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$.
Then prove that the quotient group $G/N$ is also an abelian group.
Proof.
Each element of $G/N$ is a coset $aN$ for some $a\in G$.
Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in […]

Pullback Group of Two Group Homomorphisms into a Group
Let $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms.
Define the subset $M$ of $G_1 \times G_2$ to be
\[M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.\]
Prove that $M$ is a subgroup of $G_1 \times G_2$.
[…]

If Every Nonidentity Element of a Group has Order 2, then it’s an Abelian Group
Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$.
Then show that $G$ is an abelian group.
Proof.
Let $x$ and $y$ be elements of $G$. Then we have
\[1=(xy)^2=(xy)(xy).\]
Multiplying the equality by $yx$ from the right, we […]

Elements of Finite Order of an Abelian Group form a Subgroup
Let $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$ of finite order. That is,
\[H=\{ a\in G \mid \text{the order of $a$ is finite}\}.\]
Prove that $H$ is a subgroup of $G$.
Proof.
Note that the identity element $e$ of […]

If the Order of a Group is Even, then the Number of Elements of Order 2 is Odd
Prove that if $G$ is a finite group of even order, then the number of elements of $G$ of order $2$ is odd.
Proof.
First observe that for $g\in G$,
\[g^2=e \iff g=g^{-1},\]
where $e$ is the identity element of $G$.
Thus, the identity element $e$ and the […]

Normal Subgroups, Isomorphic Quotients, But Not Isomorphic
Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$.
Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.
Proof.
We give a […]

Commutator Subgroup and Abelian Quotient Group
Let $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$.
Let $N$ be a subgroup of $G$.
Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.
Definitions.
Recall that for any $a, b \in G$, the […]

Two Quotients Groups are Abelian then Intersection Quotient is Abelian
Let $K, N$ be normal subgroups of a group $G$. Suppose that the quotient groups $G/K$ and $G/N$ are both abelian groups.
Then show that the group
\[G/(K \cap N)\]
is also an abelian group.
Hint.
We use the following fact to prove the problem.
Lemma: For a […]