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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

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

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  • $\begingroup$ And I'm wondering why Rotman would say otherwise. $\endgroup$ Apr 25 '19 at 1:18
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    $\begingroup$ You didn't explicitly answer the question "yes" or "no" (it is conventional to do so at the start of the answer). $\endgroup$ Apr 25 '19 at 3:01
  • $\begingroup$ @BillDubuque I've added this, thanks for catching that. $\endgroup$ Apr 25 '19 at 4:08
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    $\begingroup$ @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. $\endgroup$ Apr 25 '19 at 4:17

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