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.

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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 […]

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Before the proof, let us recall Lagrange's Theorem.
Lagrange's Theorem
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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 […]

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Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$.
(a) Prove that $N=\{a\in G \mid a^n=e\}$.
(b) Prove that $N=\{b^m \mid b\in G\}$.
Proof.
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Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.
Definition.
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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 […]