Problem with Cosets in the proving of $K/H \trianglelefteq G/H$

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We have $H \trianglelefteq G$ and $H \subseteq K \leq G$. we can prove if $K \trianglelefteq G$ then $K/H \trianglelefteq G/H$ like this:

We already showed that $K/H \leq G/H$. If we show that $\forall Hg \in G/H$ and $\forall Hx \in K/H$ : $(Hg)^{-1}(Hx)(Hg) \in K/H$ then the proof is complete. we have that $Hg \in G/H \Rightarrow g \in G$ and $Hx \in K/H = \{Ht | t \in K\} \Rightarrow \exists t \in K$ such that $Hx = Ht$. $\begin{align*} g \in G, t \in K & \ \ \Longrightarrow \ \ g^{-1}tg \in K \\ & \ \ \Longrightarrow \ \ H(g^{-1}tg) \in K/H \\ & \ \ \Longrightarrow \ \ (Hg^{-1}) (Ht)(Hg) \in K/H\\ & \ \ \Longrightarrow \ \ (Hg)^{-1} (Ht)(Hg) \in K/H\\ \end{align*}$

I can't undestand this part of proof

$Hx \in K/H = \{Ht | t \in K\} \Rightarrow \exists t \in K$ such that $Hx = Ht$

why we use $t$ ? can't we continue with $x$ and say $Hx \in K/H \Rightarrow x \in K$ ?

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There is a typo. $x\in K/H$ so it should say $\exists t\in K\;(x=H t)$.