Show that for the coset action $G \times G/H \to G/H$ the stabilizer of a coset is the conjugate of $H$ by $a$ Problem Statement:

Let $G$ be a group and $H<G$ a subgroup. Consider the coset action $G \times G/H \to G/H$ given by $g \cdot (aH) = gaH$ for all $g,a \in G$. prove that $$\text{stab}_G(aH) = aHa^{-1}$$


For this, the elements of the stabilizer of $aH$ by $G$ are those elements $g \in G$ such that $g \cdot aH = aH$. So, we want to show that any such element $g$ must have the form $aHa^{-1}$? If we directly substitute $g = aha^{-1}$ for any $h \in H$ we find $$aha^{-1} \cdot aH = aha^{-1}aH = aheH = ahH = aH$$ Since $h$ was an arbitrary element of $H$ we conclude that $$\text{stab}_G(aH) = aHa^{-1}$$
Is this correct? I feel as though it isn't, since the method of direct substitution seems like a bit of a brutish one but it's the only thing I could think of just to get started. Any help here is appreciated.
 A: Both of your calculations for each inclusion, the one in the question and the one in your comment are correct. To answer your question regarding "substitution", what you did was correct, it could just have been formulated more cleanly. You wanted to show $aHa^{-1} \subset \text{stab}_G(aH)$. So simply start: Let $g \in aHa^{-1}$. This means there exists $h \in H$ such that $g = aha^{-1}$. Then by the same calculation you did, this arbitrary $g$ in $aHa^{-1}$ is also in $\text{stab}_G(aH)$. The calculation for your other inclusion $\text{stab}_G(aH) \subset aHa^{-1}$ works and is interesting, since you essentially directly manipulated the coset equality $kaH = aH$ by multiplying with $a^{-1}$ from the left.
An alternative way to do this is to use the fact that for $a, b \in G$ you have $aH = bH \iff$ there exists $h \in H$ such that $a = bh$. Hence, $k \in \text{stab}_G(aH) \iff kaH = aH \iff$ there exists $h \in H$ such that $ka = ah \iff k = aha^{-1} \iff k \in aHa^{-1}$. Hence, with this argument and all the if and only ifs, both inclusions would have been shown at once, and you can really see why the coset equalities hold (because of that condition $aH = bH \iff$ there exists $h \in H$ such that $a = bh$).
