For any finite abelian group $G$, there is an integer $m$ with $G$ isomorphic to a subgroup of $U(\mathbb{Z}_{m})$. I want to prove if the following assertion from Rotmans Advanced Algebra page 205 is true:
For any finite abelian group $G$, there is some integer $m$ with $G$  isomorphic to a subgroup of $U(\mathbb{Z}_{m})$, where $U(\mathbb{Z}_{m})$ are the units of Integers module m.
Is this sentence true or false? The book I'm studying says this is not true, but I cannot find a proper counterexample to understand the mentioned claim. Thanks
 A: This is, in fact, a true statement.
There are two essential facts that I use here without proof.


*

*Any finite abelian group is a product of cyclic groups. Look up the structure theorem for modules over a PID for more on this.

*For any $ n \geq 2 $ there are infinitely many primes $p$ such that $ p \equiv 1 \text{ mod } n $. This is a special case of Dirchlet's theorem on primes in an arithmetic progression.
Now, take such a finite abelian group $ G $. We have $ G = \prod_{i=1}^n \mathbb Z/n_i \mathbb Z$, some integers $n_i$. Let $p_i$ be a prime such that $p_i \equiv 1 \text{ mod } n_i$. This exists by the second fact listed above. In fact, as there are infinitely many such primes we can take these $p_i$ to be distinct. Now, as $p_i \equiv 1 \text{ mod } n_i$, we have $ n | p_i - 1 $. Thus, we have $\mathbb Z / n_i \mathbb Z \subseteq \mathbb Z/(p_i - 1) \mathbb Z$. Strictly speaking, this means that there is an injective homomorphism between these groups, but identitying a group with its isomorphic image doesn't affect us in this case. As $p_i$ is prime, we know that $\mathbb Z / (p_i-1) \mathbb Z = (\mathbb Z/p_i \mathbb Z)^{\times}$. Hence, we have $\mathbb Z/ n_i \mathbb Z \subseteq (\mathbb Z/ p_i \mathbb Z)^{\times}$ Thus, we have $ G = \prod_{i=1}^n \mathbb Z/n_i \mathbb Z \subseteq \prod_{i=1}^n (\mathbb Z/p_i \mathbb Z)^{\times}$. One can show easily that the unit group of the product of rings is the product of their unit groups, so we have $\prod_{i=1}^n (\mathbb Z/p_i \mathbb Z)^{\times} = \left(\prod_{i=1}^n \mathbb Z/p_i \mathbb Z\right)^{\times}$. We took the $p_i$ to be distinct, hence they are pairwise relatively prime. By the Chinese Remainder Theorem, $\prod_{i=1}^n \mathbb Z/p_i \mathbb Z = \mathbb Z/ p_1 p_2 \dots p_n \mathbb Z$. Thus, $G \subseteq \left(\prod_{i=1}^n \mathbb Z/p_i \mathbb Z\right)^{\times} = (\mathbb Z/p_1 p_2 \dots p_n \mathbb Z)^{\times}$.
Field theory is listed as a tag for this, so I should mention that once you have that $\mathbb Q(\zeta_n)$ has Galois group $(\mathbb Z/n \mathbb Z)^{\times}$, this result proves the inverse Galois problem for finite abelian groups, i.e. all finite abelian groups are Galois groups over $\mathbb Q$.
