Let $t\in\Bbb N,t\ge 1,(G,+,0)$ abelian group $|G|=n$ and $\gcd(t,n)=1$ Prove $f(x)=tx$ is an automorphism

Question

Let $t\in\mathbb{N}, t\geq 1,\, (G,+,0)$ abelian group $|G|=n$ and $\gcd(t,n)=1$

Let $f(x)=tx=\underbrace{x+\cdots+x}_{t\text{ summands}}.$

Prove $f\in \text{Aut}(G)$

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I proved $f$ is an homomorphism now I want to prove $f$ is an isomorphism

• $f$ is injective because if $f(x)=f(y)$ $$tx=ty \Longleftrightarrow tx-ty=0 \Longleftrightarrow t(x-y)=0 \Longleftrightarrow x-y=0 \Longleftrightarrow x=y $$
• $f$ is surjective?

I think it is because if $y\in f(G)$ then $y=tx$ for $x\in G$ then $$f(x)=tx=y$$

Is this correct? I never used $\gcd(n,t)=1$ so why is that important?

Answer 1

Hint: Since $\gcd(n,t)=1$, by Bézout's Theorem, there exist $x,y\in\Bbb Z$ such that

$$xt+yn=1.$$

Consider

$$\begin{align} g:G&\to G,\\ a&\mapsto xa. \end{align}$$

Answer 2

Your proof of injectivity is incorrect. Think about what injectivity means: it means that if $x_1$ and $x_2$ are two elements of $G$, and $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Your proof of surjectivity is also incorrect. Think about what surjectivity means: it means for every $y \in G$, you need to SHOW there is some $x \in G$ such that $y =f(x)$.

Yes, $\operatorname{GCD}(t,n) = 1$ is important and you need it in the proof of injectivity.