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.

(a) The additive group of rational numbers $\Q$ is divisible.

We know that $\Q$ is a nontrivial abelian group.
Let $a\in Q$ and $k$ be a nonzero integer.
Since $\Q$ is an additive group, we are seeking $x\in \Q$ such that
\[kx=a.\]

Since $k$ is a nonzero integer, we have a solution $x=a/k\in \Q$.
Thus $\Q$ is divisible.

(b) No finite abelian group id divisible.

Let $G$ be a finite abelian group of order $|G|=n$.
If $G$ is trivial, that is, $n=1$, then by definition, $G$ is not divisible.
So let us assume that $G$ is nontrivial.

We claim that $G$ is not divisible since there is no $n$-th root of a nonidentity element of $G$.
Let $a\in G$ be a nonidentity element of $G$.
(Such an element exists because $G$ is nontrivial.)

If there is $x\in G$ such that
\[x^n=a,\]
then by Lagrange’s theorem we have $x^n=e$, the identity element of $G$.
This implies that $a=e$, and this contradicts our choice of $a$.
Thus $G$ is not divisible.

Group of Order $pq$ is Either Abelian or the Center is Trivial
Let $G$ be a group of order $|G|=pq$, where $p$ and $q$ are (not necessarily distinct) prime numbers.
Then show that $G$ is either abelian group or the center $Z(G)=1$.
Hint.
Use the result of the problem "If the Quotient by the Center is Cyclic, then the Group is […]

Use Lagrange’s Theorem to Prove Fermat’s Little Theorem
Use Lagrange's Theorem in the multiplicative group $(\Zmod{p})^{\times}$ to prove Fermat's Little Theorem: if $p$ is a prime number then $a^p \equiv a \pmod p$ for all $a \in \Z$.
Before the proof, let us recall Lagrange's Theorem.
Lagrange's Theorem
If $G$ is a […]

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

A Simple Abelian Group if and only if the Order is a Prime Number
Let $G$ be a group. (Do not assume that $G$ is a finite group.)
Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.
Definition.
A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]

A Subgroup of the Smallest Prime Divisor Index of a Group is Normal
Let $G$ be a finite group of order $n$ and suppose that $p$ is the smallest prime number dividing $n$.
Then prove that any subgroup of index $p$ is a normal subgroup of $G$.
Hint.
Consider the action of the group $G$ on the left cosets $G/H$ by left […]

Nontrivial Action of a Simple Group on a Finite Set
Let $G$ be a simple group and let $X$ be a finite set.
Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.
Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.
Proof.
Since $G$ acts on $X$, it […]

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 […]

The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$
Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.
Proof.
Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]