Decomposition of $\mathrm{Aut}(\mathbb{Z}_n)$

Question

I know that $\mathrm{Aut}(\mathbb{Z}_n)\cong \mathbb{Z}_n^\times$ and in this thread a user correctly states that $\mathbb{Z}_n\cong \mathbb{Z}_{p_1}\times\dots\times \mathbb{Z}_{p_k}$, where the $p_i$ are $n$'s prime decomposition, and also $\mathbb{Z}_p^\times\cong \mathbb{Z}_{p-1}$.

In my case, I have a theorem stating that there are unique $1<d_1\,|\,d_2\,|\,\dots\,|\,d_r$ such that $G\cong \mathbb{Z}_{d_1}\times\dots\times \mathbb{Z}_{d_r}$. The problem is the strict inequality $d_1>1$. I want to decompose $\mathbb{Z}_{400}^\times$, but with the method above, I only get $\mathbb{Z}_{400}^\times\cong (\mathbb{Z}_1)^5\times (\mathbb{Z}_4)^2$ and clearly $1\ngtr1$.

Answer 1

Let us ask the computer, to have the answer, then we will understand it humanly:

sage: A = IntegerModRing(400)
sage: U = A.unit_group()
sage: U
Multiplicative Abelian group isomorphic to C2 x C4 x C20

So the theorem of invariant factors works for the unit group $U=A^\times$ of the abelian group $A=(\Bbb Z/400, +)$, seen as underlying abelian group of the ring $(\Bbb Z/400, +,\cdot)$. The elements of $U$ are of the shape $k$ for $k$ from $1$ to $399$ (taken modulo $400$) which is relatively prime to $400$. There are $$ \varphi(400) =400\left(1-\frac 12\right)\left(1-\frac 15\right) = 400\cdot \frac 12\cdot\frac45 =160 $$ such elements. The invariant factors are $d_1=2$, $d_2=4$, $d_3=20$, with $1$ divides $d_1=2$, which divides $d_2=4$, which divides $d_3=20$. Explicitly: $$ \color{blue}{ \boxed{\ (\Bbb Z/400)^\times \cong (\Bbb Z/2)\times (\Bbb Z/4)\times (\Bbb Z/20)\ . }} $$

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A way to see this result is as follows.

The ring $\Bbb Z/400$ can be written isomorphically as the product of rings $(\Bbb Z/2^4)\times(\Bbb Z/5^2)$. The units group of this ring corresponds to the multiplicative group $$ (\Bbb Z/2^4)^\times \times(\Bbb Z/5^2)^\times\ . $$ So let us understand the structure of the two groups explicitly.

Again, the computer tells us...

sage: IntegerModRing(16).unit_group()
Multiplicative Abelian group isomorphic to C2 x C4
sage: IntegerModRing(25).unit_group()
Multiplicative Abelian group isomorphic to C20

(The prime two is the oddest one for such computations.) So we will show:

• The multiplicative group $(\Bbb Z/16)^\times$ is isomorphic to the additive group $(\Bbb Z/2)\times(\Bbb Z/4)$.
• The multiplicative group $(\Bbb Z/25)^\times$ is isomorphic to the additive group $\Bbb Z/20\cong\Bbb Z/4\times \Bbb Z/5$.

For the second isomorphism, it is enough to exhibit in $(\Bbb Z/25)^\times$ two elements, one of order four, one of order five. An element of order four is $7$, since $7^2=49=-1$ is not one, but $7^4$ it is. An element of order five is $6$, since $6^5=7776=1$.

For the first isomorphism we know that the isomorphic group has $\varphi(16)=8\left(1-\frac1 2\right)=8$ elements. Let us search first for the elements of order two in $(\Bbb Z/16)^\times$. So we solve $x^2=1$ modulo $16$. It turns out that there are four solutions, here are they: $0\pm 1$, $8\pm 1$. Three of them, $-1,7,9$ have order two. So we have two invariant factors, and these correspond to the only one possible decomposition $(\Bbb Z/16)^\times\cong(\Bbb Z/2)\times(\Bbb Z/4)$. (The elements of order four are $4\pm1$, $12\pm 1$, $18\pm 1$. There are no elements of order eight.)