Show this map is a permutation

Question

I'm trying to show that the map $\{ 0,\ldots,n-1\} \rightarrow \{0,\ldots,n-1\}$, defined by $$x\mapsto x+a \text{ mod} n$$ belongs to $\mathcal{S_n}$,where $g.c.d(a,n)=1$ and I need a check.

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To this aim, I want to show that this is a bijection:

• injectivity: $f(i)=f(j)$ implies $$(i+a) \text{mod}n = (j+a)\text{mod n}$$ from which follows $$n|(i-j)$$ and hence $i-j$ is a multiple of $n$, therefore they're equal mod n. This proves injectivity

• surjectivity: I have to show that for every $j \in \{ 0,\ldots,n\}$ there exists $i \in \{0,\ldots,n-1\}$ such that $f(a)=i$, which means $$ (i+a)\text{ mod n} = j $$

which means that $$i+a = q*n + j$$ for some $q$ and hence $$i = j -a \text{ mod n}$$

Therefore I determined $i \in \{0,\ldots,n-1\}$ and the proof is complete.

Is everything okay?

Answer 1

Yes, it is correct. But a function from a finite set into itself is injective if and only it is surjective. So, proving injectivity is enough.

Answer 2

The map is invertible, the inverse being $y\mapsto y-a$.

Invertible maps are bijective.