In Hungerford's Abstract Algebra, they state the following Corollary:
"Let $N$ be a normal subgroup of a group $G$ and let $K$ be any subgroup of $G$ that contains N. Then $K$ is normal in $G$ $\iff$ $K/N$ is normal in $G/N$."
In the proof of this corollary, they state that if $a\in G$ and $k\in K$, then the following equation follows from the normality of $K/N$: $$\tag{1} N a^{-1} k a=\left(N a^{-1}\right)(N k)(N a)=(N a)^{-1}(N k)(N a) \in K / N $$ My question is where exactly the "normality" comes in to the picture here? $N a^{-1} k a=\left(N a^{-1}\right)(N k)(N a)$ follows from the definition of the binary operation in the quotient set. I would then assume that $Na^{-1}=(Na)^{-1}$ should follow from normality. But consider $$\tag{2} N a \cdot N a^{-1}=N a a^{-1}=N e=N a^{-1} a=N a^{-1} \cdot N a. $$ This shows that $Na^{-1}$ is the inverse of $Na$ without using normality, so why exactly does eq. $(1)$ require this condition?
Question is, why does $(Na)^{-1}(Nk)(Na)$ belong to $K/N$? That's not a product of $3$ elements of $K/N$, we just conjugated an element of $K/N$ by an element of $G/N$. But that's exactly where the assumption $K/N\trianglelefteq G/N$ is used, it implies that this conjugate indeed belongs to $K/N$.