# Fundamental Theorem of Finite Abelian Groups and the mulitplicative group of a field.

(a) State the Fundamental Theorem of Finite Abelian Groups (invariant version).

(b) Consider the group $K^{\times}$, under multiplication, of all non-zero elements in a field $K$. Let $A$ be a finite subgroup of $K^{\times}$ of order n. Determine the invariant factors of $A$.

(a) From Advanced Modern Algebra (Rotman):

Every finite abelian group $G$ is a direct sum of cyclic groups, $$G=J(c_1) \oplus ... \oplus J(c_r),$$

where $r \geq 1$, $J(c_i)$ is a cyclic group of order $c_i$, and $$c_1 | c_2 |...|c_r.$$

(b) From Basic Abstract Algebra (Robert Ash):

The multiplicative group of a finite field is cyclic. More generally, if G is a finite subgroup of the multiplicative group of an arbitrary field, then $G$ is cyclic.

So we know that $K^{\times} \cong \Bbb{Z}_n$ for some $n$. So we just find the invariant factors for $\Bbb{Z}_n$ like we do for any finite abelian group. We write it in the form $\Bbb{Z}_n \cong G_{p_1} \oplus G_{p_k}$, where $G_{p_i}$ is the $p_i$-primary subgroup. And then we find $c_1, ... ,c_r$ and we are done.

My Question:

I'm not really sure if I understood the question correctly, because isn't the answer obvious? We can find the invariant factors of $K^{\times}$ like we do for any group. I'm not really sure if I understood what the question is really asking for.

• Note that $K$ may be an infinite field, so you are to find the invariant factors of $A\subseteq K^\times$, which is a finite subgroup. But besides that, I do not see whether there is anything deeper either. – Karl Kronenfeld Aug 20 '13 at 0:18
• I think fact (b) is often proven using the Fundamental Theorem of Finite Abelian Groups (see Proposition 18 in section 9.5 of Dummit and Foote, for instance). So maybe that is what the question is getting at. – user55407 Aug 20 '13 at 3:10

The question wants you to explicitly prove (b) from (a). Notice the only invariant factor of a cyclic group $\cong \Bbb Z/a_1\Bbb Z$ is just $\{a_1\}$ (a singleton set clearly satisfies the requirements to be an invariant factor decomposition).
Hint: To show $A$ is cyclic consider the maximum number of solutions to the equation $x^k=1$ in any field and look at any non-cyclic invariant factor decomposition.