Let $\sigma$ : $\mathbb{Z}_{11} \to \mathbb{Z}_{11}$ be given by $\sigma([a]) = [5a + 3]$. Prove that $\sigma$ is bijective.
Approach It has to be one to one and onto so
It is one to one if $\sigma([a1])=\sigma([a2])$
$[5a1+3]=[5a2+3]$
$[5a1]=[5a2]$
$[a1]=[a2]$
Does this approach work?. How to show that it's onto
it would be like $\sigma([y-3/5])=[5(y-3)/5+3]$.
On
It is a basic fact of discrete mathematics that for a finite set $A$ an injective map $f:\>A\to A$ is automatically bijective. Therefore it is sufficient to prove injectivity of your $\sigma$. Now $\sigma([a])=\sigma([b])$ can be unpacked to $$(5a+3)=(5b+3 )\ + 11k, \qquad k\in{\mathbb Z}\ .$$ This implies $11\> |\> 5(a-b)$, hence $11\> |\> (a-b)$. The latter is saying that $[a]=[b]$.
To show that $\sigma$ is surjective (or "onto"), we have to show that $\forall y \in Y, \exists x \in X \text{ such that } y = \sigma(x)$ (see. this Wikipedia site). Here $Y = Z_{11} = X$. Because there is no further information about $Z_{11}$ this might not be fulfilled. Example: $Z_{11} = \{ 8 \}$ this would mean that for $y = 8 \in Y = \{ 8 \}$ there must exist an element $x \in X= \{ 8 \}$ such that $8 = 5x + 3$. But the $x$ solving this equation is $1$ which is not in $ X= \{ 8 \}$.