What is the difference between $[H, g]$ and $[h, g]$?

Question

I am working on this problem, where $[H, g]$ is the commutator group:

Let $H$ be a subgroup of $G$, show that $[H, g] = [H, \langle g \rangle]$.

Before solving it, I need to understand the difference between $[H, g]$ and $[h, g]$ because the writer of this problem used $[H, g]$ not without a purpose. Therefore here is my question:

What is the difference between the set builders of $[H, g]$ and $[h, g]$ in term of solving this problem?

Thank you for your time.

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I use the set builder given by @nullUser but am still stuck one step short:

$$\begin{align} [H,g] :&= \{[h,g] : h \in H \} = \{hgh^{-1}g^{-1} : h \in H \}\\ &= \{hg(g^{n-1})(g^{n-1})^{-1}h^{-1}g^{-1} : h \in H \} \\ &= \{h(gg^{n-1})(g^{n-1})^{-1}h^{-1}g^{-1} : h \in H \} \\ &= \{hg^n\underbrace{(g^{n-1})^{-1}h^{-1}}_{\text{non-commutative!}}g^{-1} : h \in H \} \\ \end{align}$$

If only the elements are commutative then I am done. This was the reason behind my posting: Perhaps I have been so carried away by the usual set builder of $[h, g]$ that I missed something from $[H, g]$. Any other ideas of proving it?

Answer 1

$[H,g]$ denotes the set of all commutators of the form $[h,g]$ with $h \in H$

Answer 2

Since you want set builder notation:

$$ [H,g] := \langle\{[h,g] : h \in H \}\rangle = \langle\{hgh^{-1}g^{-1} : h \in H \}\rangle $$

where here the angled brackets indicate that we need to take the subgroup generated by elements of this form. (Thanks to comments for pointing this out.)