I understand how any subgroups are cyclic and there is a subgroup for each divisor of $d$ Let's say for example
Let $d$ be a divisor of $n=|G|$. Consider $H=\{ x \in G : x^d =e\}$. Then $H$ is a subgroup of $G$ and $H$ contains all elements of $G$ that have order $d$ and $d$ is equal to 4
can anyone explain?
Many ways to answer that, based on how much group theory you've done so far.
The most elementary way I can think of to compute it is using the following theorem, which is not hard to prove: A cyclic group G of finite order has the following property: For each d which divides |G|, there is a unique subgroup (which is cyclic) of order d. This in fact is if and only if, but we won't need that.
With that theorem your question becomes easy. The only elements $x$ such that $x^4=e$ are elements of order 1, 2, 4. There is a unique element of order 1, the identity element. Given that there is a unique subgroup of order 2, that means there is a unique element of order 2, say $g$, which forms the subgroup of order 2
$$ H=\{g, g^2=e\} $$
Also there is a unique cyclic subgroup K of order 4, say $K=\langle h\rangle =\{h, h^2, h^3, h^4=e \}$. You can immediately see that $h^2$ has order 2 so it must be equal to g because of the uniqueness we discussed above. Also, $h^3$ has order 4. Therefore we have 4 elements with order up to 4, namely $e, g, h, h^3$. We know that there are no more such elements because that would contradict the above theorem, as any new element, say $x$ of order either 2 or 4, distinct from the above, would generate a new subgroup of order 2 or 4, namely $N=\langle x\rangle$.
