# 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. Add to solve later

## Proof.

### (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. Add to solve later

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