Let $G$ be a group and $K$ be a subgroup of $G$. The following definitions of normality are equivalent:
(i): $K \triangleleft G \iff$ for all $g \in G, k \in K$, $gkg^{-1} \in K$.
(ii): $K \triangleleft G \iff$ for all $g \in G, gK = Kg$.
The lecture notes I'm using use a fairly convoluted proof of (i) $\implies$ (ii), which I'm struggling to follow. I came up with another proof that seems far too simple, so I'm wondering if there's a flaw in it. Here it is:
Assume that for all $g \in G, k \in K, gkg^{-1} \in K$. $gkg^{-1} \in K \implies gk \in Kg$, so $gk \in Kg$ for all $k \in K, g \in G$ i.e. $gK \subset Kg$ for all $g \in G$.
Also for all $g \in G, g^{-1} \in G$, so repeat the same argument but substituting $g$ with $g^{-1}$ in the premise to give that $Kg \subset gK$. This completes the proof.
Is this valid? I am particularly worried about the statement $gkg^{-1} \in K \implies gk \in Kg$. This seems obvious to me but perhaps it's not sufficiently rigorous in a proof.