Error in claiming $N_0$ = $G_0$ when $N/N_0 \leq G/G_0$?

Question

Let $N \leq G, N_0 \unlhd N, G_0 \unlhd G$ such that $N/N_0 \leq G/G_0$. Since the identity elements in $N/N_0$ and $G/G_0$ are same, $N_0$ = $G_0$. What is the error in this claim? It would be helpful if a concrete explanation is given. Plus a example would be ideal. When we say $N/N_0 \leq G/G_0$, what do we actually mean? Is it that there needs to exist only an isomorphic copy of $N/N_0$ in $G/G_0$?

Answer 1

Strictly speaking the claim is correct: the elements of $N/N_0$ are cosets for $N_0$, and saying that these are also in $G/G_0$, which implies they are cosets for $G_0$, is only possible if $N_0=G_0$.

One can however give a weaker sense to the statement $N/N_0\leq G/G_0$, namely that the inclusion $N\hookrightarrow G$ induces an injective morphism $N/N_0\to G/G_0$. As far as I can see the conditions given do not imply that the inclusion induces any morphism at all; the condition that would make such an induced morphism exist is $N_0\leq G_0$ (it is then the composition of the natural maps $N/N_0\to N/G_0\to G/G_0$). If one assumes that, the induced morphism is injective if and only if $N_0=N\cap G_0$, and this is of course weaker than $N_0=G_0$.