Normal subgroups are not transitive, but $H \subseteq N \lhd G, H \lhd G \implies H \lhd N$?

Question

Let $H,N,G$ be groups where $H \subseteq N \subseteq G$. For the 3 statements below:

1. $H$ is a normal subgroup of $G$: $ghg^{-1} \in H \ \forall h \in H, \forall g \in G$

2. $N$ is a normal subgroup of $G$: $gng^{-1} \in N \ \forall n \in n, \forall g \in G$

3. $H$ is a normal subgroup of $N$: $nhn^{-1} \in H \ \forall h \in H, \forall n \in N$

Is 1 enough to imply 3?

Answer 1

I think we just let $g=n$:

If $H$ is a normal subgroup of $G$, then $ghg^{-1} \in H \ \forall h \in H, \forall g \in G$ including $n \in N \subseteq G$

Therefore, not only

$H \subseteq N \lhd G, H \lhd G \implies H \lhd N$

but also

$H \subseteq N \subset G, H \lhd G \implies H \lhd N$?