According to Approach0, this is a new question to MSE.

The Details:

Definition: The group $U(n)$ of units modulo $n$ is the group under multiplication modulo $n$ of all equivalence classes $[a]_n$ modulo $n$ such that $\gcd(n, a)=1.$

Lemma: $$U(n)\cong {\rm Aut}(\Bbb Z_n).$$

For a proof, see Theorem 6.5 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)."

The Question:

Is there a standard/elegant group presentation for $U(n)$?

My Attempt:

Thinking that I could use the Lemma above, the closest thing I could find in the literature is a result by Baumslag (1963), stated as Theorem 4.8 of Lyndon & Schupp's "Combinatorial Group Theory":

Theorem: If $G$ is a finitely generated residually finite group, then ${\rm Aut}(G)$ is also residually finite.

But both $\Bbb Z_n$ and $U(n)$ are trivially finite, so this doesn't tell us anything new, since all finite groups are residually finite. (See the second section of the Wikipedia article linked to above.)

Come to think of it (some more), the theorem is completely useless here, but I've typed it up now; besides, it really is the closest thing I could find.

My Google searches were along the lines of "presentation for the group of units modulo n". Nothing useful showed up.

Perhaps it's something easy that I've overlooked.

Please help :)


Not sure whether this is standard or whether you would call that elegant (probably not), but the thing that comes first to my mind (and what I would do first) is the following:

Let $n = p_1^{\nu_1} \cdots p_r^{\nu_r}$ be the prime decomposition of $n$. Then we get $$\mathbb{Z}/n\mathbb{Z}^{\times} \cong \mathbb{Z}/p_1^{\nu_1}\mathbb{Z}^{\times} \times \dots \times \mathbb{Z}/p_r^{\nu_r}\mathbb{Z}^{\times}$$ by the Chinese remainder theorem and additionally we have

$\mathbb{Z}/p^{\nu}\mathbb{Z}^{\times} \cong \begin{cases} C_1 & p = 2 \;\text{and}\; \nu = 1\\ C_2 \times C_{2^{\nu -2}} & p = 2 \;\text{and}\; \nu \geq 2\\ C_{\varphi({p^\nu})} = C_{p^{\nu - 1}(p - 1)} & \text{otherwise} \end{cases}$

Now if you have two groups $G = \langle S \mid R \rangle$ and $H = \langle S' \mid R' \rangle$, then $$G\times H\cong \langle S\cup S'\mid R\cup R'\cup\{ss'=s's\mid s\in S, s'\in S'\}\rangle.$$ Thus you can give a presentation of $\mathbb{Z}/n\mathbb{Z}^{\times}$ depending on the prime decomposition, where the easiest case is the one that the highest power of $2$ dividing $n$ is $2$ as you only need to combine presentations of finite cyclic groups then.

  • $\begingroup$ But $$G\times H\cong \langle S\cup S'\mid R\cup R'\cup\{ss'=s's\mid s\in S, s'\in S'\}\rangle.$$ $\endgroup$ – Shaun Oct 4 at 22:12
  • 1
    $\begingroup$ Yes, you are right. Just saw that and wanted to change that as well haha. $\endgroup$ – ThorWittich Oct 4 at 22:13

These groups are usually cyclic. In particular, if $8$ does not divide $n$ then the group is cyclic: $$ \langle a\mid a^{\phi(n)}=1\rangle $$

Otherwise (where $8$ divides $n$), we get $$ \langle a,b\mid a^{\phi(n)/2}=b^2=[a,b]=1\rangle $$

  • $\begingroup$ I don't understand why the second presentation holds. Please would you elaborate? $\endgroup$ – Shaun Oct 4 at 22:37

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