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Let $f:\mathbb{Z}_n\rightarrow \mathbb{Z}_n$ a group homomorphism. Prove $f\in\text{Aut}(\mathbb{Z}_n)\Longleftrightarrow f([1]_n)\in \mathcal{U}(\mathbb{Z}_n)$ $\mathcal{U}(\mathbb{Z})_n:=\{[a]_m\in\mathbb{Z}_m|\gcd(a,m)=1\}$

My proof

$\Rightarrow$

$\forall [x]\in \mathbb{Z}_n $ $$f([x])=f(x[1])=xf([1])$$

So can I say $f([1])$ is a generator and then $f([1])\in\mathcal{U}(\mathbb{Z}_n)$ ?

$\Leftarrow $

  • $f$ is injective: for $[x],[y]\in \mathbb{Z}_n$ such that $f([x])=f([y])$ $$f([x])=f([y])$$ $$f(x[1])=f(y[1])$$ $$xf([1])=yf([1])$$ $$x=y$$ $$[x]=[y]$$

  • $f$ is surjective: for $[y]\in \mathbb{Z}_n$ exists $[x]=[yf^{-1}(1)]$ such that $$f([x])=f([yf^{-1}(1)])=yf([f^{-1}(1)])=y$$

Is my proof correct?

Shaun
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Jhon Knows
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    @Nightflight I guess $G = \Bbb Z_n = \Bbb Z/n\Bbb Z$ and $\mathcal U(\Bbb Z_n)$ means $(\Bbb Z/n\Bbb Z)^\times$. – WhatsUp Oct 31 '21 at 22:30

2 Answers2

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In $\Leftarrow$ part, you cannot claim that $f^{-1}([1])$ exists, because you don't prove $f$ is surjection.

Since $\mathbb Z_n$ is finite set with cardinality $n$, so when you prove $f:\mathbb Z_n\to\mathbb Z_n$ is injective, then $f$ is automatically an automorphism (as a group, a ring ... any algebraic structure) by the Pigeonhole Principle.

So, adding some details of $\Rightarrow$ part, my amendment based on your proof is as follows.


Proof

$\Rightarrow$ part.

Since $f$ is group automorphism, so there exists $[x]\in\mathbb Z_n$ such that $f([x])=[1]$. Let $f([1])=[a]$.

Then $f([x])=f([x][1])=[x]f([1])=[x][a]=[xa]$, so $\gcd(a,n)=1$. so $f([1])\in\mathcal U(\mathbb Z_n)$.

$\Leftarrow$ part.

Since $f([1])\in\mathcal U(\mathbb Z_n)$, there exists $[a]\in\mathbb Z_n$ such that $[a]f([1])=[1]$. So if $f([x])=f([y])$, then $[x]f([1])=[y]f([1])$. $[x]=[y]$. $f$ is injection, so bijection also.

MH.Lee
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So can I say $f([1])$ is a generator and then $f([1])\in\mathcal{U}(\mathbb{Z}_n)$?

Yes.

As for the other direction, since $\Bbb Z_n$ is finite, it suffices to show that $f$ is injective for it to be bijective. Your proof there is okay.

I'm not sure about your proof of surjectivity; luckily, by the above, it is not necessary.

Shaun
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