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My example says:

AutC$_4$, $C_4 = \{1, a, a^2, a^3\}$

And then it points to $a$ and $a^3$ and says these are generators and these are the two elements of order 4. Why is this?

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up vote 2 down vote accepted

Because only $a$ and $a^3$ generate the group. This means that you can get every element of $C_4$ by using (maybe several times) $a$ or $a^3$ only. $1$ and $a^2$ generate only a subgroup, reps. $\{1\}$ and $\{1, a^2\}$.

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a is a generator of the group as every element of the group can be expressed by a^n (n is positive, negative or zero) you can check that by yourself, also you can prove the same for a^3, now order of an element (say a) of a group is the least positive integer n such that a^n = identity element, you can check this for a and a^3 in your group, but as the group is cyclic, it's easier to tell their order is 4 because for cyclic group order of generator element = order of the group, order of the group is 4 so all generator of this group will be of order 4 also

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A nice result about $\textit{finite}$ cyclic groups is the following:

Let $G$ be a cyclic group of order $n$. Then the number of generators of $G$ is $\phi (n)$, where $\phi$ is Euler's function.

$\phi (4)=4(1 - \frac{1}{2})=2$, so $C_4$ has $2$ generators.

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