Tagged: abelian group

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.

 
Read solution

LoadingAdd to solve later

Eckmann–Hilton Argument: Group Operation is a Group Homomorphism

Problem 268

Let $G$ be a group with the identity element $e$ and suppose that we have a group homomorphism $\phi$ from the direct product $G \times G$ to $G$ satisfying
\[\phi(e, g)=g \text{ and } \phi(g, e)=g, \tag{*}\] for any $g\in G$.

Let $\mu: G\times G \to G$ be a map defined by
\[\mu(g, h)=gh.\] (That is, $\mu$ is the group operation on $G$.)

Then prove that $\phi=\mu$.
Also prove that the group $G$ is abelian.

 
Read solution

LoadingAdd to solve later

No Finite Abelian Group is Divisible

Problem 240

A nontrivial abelian group $A$ is called divisible if for each element $a\in A$ and each nonzero integer $k$, there is an element $x \in A$ such that $x^k=a$.
(Here the group operation of $A$ is written multiplicatively. In additive notation, the equation is written as $kx=a$.) That is, $A$ is divisible if each element has a $k$-th root in $A$.

(a) Prove that the additive group of rational numbers $\Q$ is divisible.

(b) Prove that no finite abelian group is divisible.

 
Read solution

LoadingAdd to solve later

Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself

Problem 221

Let $p$ be a prime number. Let
\[G=\{z\in \C \mid z^{p^n}=1\} \] be the group of $p$-power roots of $1$ in $\C$.

Show that the map $\Psi:G\to G$ mapping $z$ to $z^p$ is a surjective homomorphism.
Also deduce from this that $G$ is isomorphic to a proper quotient of $G$ itself.

 
Read solution

LoadingAdd to solve later

Group Homomorphism, Conjugate, Center, and Abelian group

Problem 209

Let $G$ be a group. We fix an element $x$ of $G$ and define a map
\[ \Psi_x: G\to G\] by mapping $g\in G$ to $xgx^{-1} \in G$.
Then prove the followings.
(a) The map $\Psi_x$ is a group homomorphism.

(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$, where $Z(G)$ is the center of the group $G$.

(c) The map $\Psi_y=\id$ for all $y\in G$ if and only if $G$ is an abelian group.

 
Read solution

LoadingAdd to solve later