# The quotient of a finitely generated group is a finite group.

Let $K$ be a number field , $\mathcal{O}_K$ the ring of integers , and $\mathcal{O}_K ^\times$ the unit group of $\mathcal{O}_K$.

Then using the Dirichlet's theorem , we know that $\mathcal{O}_K ^\times$ is finitely generated group , and so $(\mathcal{O}_K ^\times )^m$ is also finitely generated.

In this conditions , is $\mathcal{O}_K ^\times / (\mathcal{O}_K ^\times )^m$ finite group?

I'm sorry that my question is very elementary.

I think the claim holds more generally , but I don't know the proof. Thank you.

• Do you know the structure theorem for finitely generated abelian groups? [That's more than needed, it suffices to know that a finitely generated abelian torsion group is finite.] – Daniel Fischer Nov 21 '17 at 15:47
• Yes , but I don't know how to use. – GreenTea Nov 22 '17 at 0:48

Yes, under these conditions the quotient is finite. By Dirichlet's unit theorem we have $\mathcal{O}_K^\times \cong \mathbb{Z}^{r+s-1} \times F$ where $r$ is the number of embeddings $K \hookrightarrow \mathbb{R}$, $s$ is the number of conjugate pairs of embeddings $K \hookrightarrow \mathbb{C}$, and $F$ is the finite group of roots of unity in $K$.
Note, that the group law on the left-hand side is multiplication while the group law in $\mathbb{Z}^{r+s-1}$ is addition. (I am keeping it this way since it is usually easiest to think of a free abelian group as a power of $\mathbb{Z}$.)
Thus $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^m \cong \mathbb{Z}^{r+s-1} / m \mathbb{Z}^{r+s-1} \times F/F^m$. Any quotient of $F$ is finite and $\mathbb{Z}^{r+s-1} / m \mathbb{Z}^{r+s-1} \cong (\mathbb{Z}/m\mathbb{Z})^{r+s-1}$ is of order $m^{r+s-1}$ whence also finite.