Finding a stabilizer of subset of Group Let $S = \{1,2,3,4\}$ and $G = S_4$. Find $G_{1,2,3}$ and $G_4$.
So I'm solving for stabilizer of $\{1,2,3\}$ and $\{4\}$ in $G$. Two different stabilizers.
My thought is that I need to look for $g \in G$ such that $g(t) \in T$, where $T$ is $\{1,2,3\}$ or $\{4\}$.
When $T$ is $\{1,2,3\}, ~G_T$ is $\{(), (2,3), (1,2), (1,2,3), (1,3,2), (1,3)\}$ because those $g$'s make $1,~2,$ or $3$ go back to set $\{1,2,3\}$, which is $T$ itself.
Likewise, when $T$ is $\{4\}$, I only to need to choose permutations that don't include $4$. So it's $\{(), (2,3), (1,2), (1,2,3), (1,3,2), (1,3)\}$.
Is my thought correct?

This is a particular case of this more general situation: let $S$ be a finite set, $X\subset S$ and $f\in\operatorname{Sym}(S)$; then, by bijection properties, $f(X)=X\iff f(X^c)=X^c$ (here $X^c:=S\setminus X$). Therefore:
\begin{alignat}{1}
\operatorname{Stab}(X) &= \{f\in \operatorname{Sym}(S)\mid f(X)=X\} \\
&= \{f\in \operatorname{Sym}(S)\mid f(X^c)=X^c\} \\
&= \operatorname{Stab}(X^c) 
\end{alignat}
In your case: $S=\{1,2,3,4\}$ (whence $\operatorname{Sym}(S)=S_n$), $X=\{1,2,3\}$ and $X^c=\{4\}$.
