Is the following statement true about generating set of a group? Suppose $G$ be a group. Let $X$ be a subset of $G$ and $H$ be a subgroup of $G$. Then is the following statement true?
Suppose that $$G=\langle X, H\rangle.$$
Can we write $$G=\langle X\rangle H?$$ where the above product is internal product of subgroups.
My Attempt: if $H$ is trivial subgroup, then above statement is true. For non-trivial subgroup, it seems to me that it is correct. But unable to have a proof of the same. Please help.
 A: No, this is not true in general.
Consider $G=\mathbb{F}_2$ the free group on two generators $a,b\in\mathbb{F}_2$. Then let $X=\{a\}$ and $H=\langle b\rangle$. Note that clearly $G=\langle X,H\rangle$, but it is not true that $G=\langle X\rangle H$, because the set on the right is of the form $\{a^nb^m\}$, and it does not contain for example $ba$.
EDIT: Note that the statement is true when at least one of $H$, $\langle X\rangle$ is normal. More generally, given two subgroups $H$, $K$ if one of them is normal, then $HK=KH$. And this implies that $\langle H,K\rangle=HK$.
A: May be this is an overkill, but I'd like to provide the next example.
It is known that a group generated by $2$ involutions is dihedral. If your statement is true, then $|G|=|\langle X\rangle H|=\dfrac{|\langle X\rangle||H|}{|\langle X\rangle \cap H|}=|\langle X\rangle||H|=4$, for every dihedral group $G$, which is not a case, of course.
It is worth to note that it is true, when $H$ is the Frattini subgroup $\Phi(G)$, that can be characterized as the set of all non-generating elements.
