# 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

• What book are you studying? Apr 24 '19 at 20:58
• What is $U(\Bbb Z_m)$? Apr 24 '19 at 21:23
• units in the ring of integers modulo $m$ ? Apr 24 '19 at 21:27
• Already edited. I'm studying Rotmans Advanced Algebra and U(Zm) means units of Integers module m
– Cos
Apr 24 '19 at 22:17
• Probably follows from the Kronecker-Weber Theorem and this. Apr 24 '19 at 22:39

This is, in fact, a true statement.

There are two essential facts that I use here without proof.

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

2. 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 $$\mathbb Q$$.

• And I'm wondering why Rotman would say otherwise. Apr 25 '19 at 1:18
• You didn't explicitly answer the question "yes" or "no" (it is conventional to do so at the start of the answer). Apr 25 '19 at 3:01
• @BillDubuque I've added this, thanks for catching that. Apr 25 '19 at 4:08
• @QuangHoang In the copy of Rotman I have, the remark on page 205 says "We cannot conclude more from the proposition; given any finite abelian group G, there is some integer m with G isomorphic to a subgroup of U(Im)". It seems to me that Rotman is saying that the result is true, but it isn't a corollary of the result he just proved (that $\text{Gal}(\mathbb F(\zeta_n)/F)$ is a subgroup of $(\mathbb Z/m \mathbb Z)^{\times})$. The phrasing is kind of poor, so I see where the confusion arises. Apr 25 '19 at 4:17