Let $G$ be a group and $H$ be a proper subgroup of $G$. Let $N$ be a normal subgroup of both $G$ and $H$.
Question: Is it possible that $G/N = H/N$?
Motivation: For a finite extension $L/K$ of local fields, we have exact sequences
$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} 1& \ra{} & I_L & \ra{} & G_L & \ra{} & \operatorname{Gal}(L^{ur}/L) & \ra{} & 1 & & \\ & & \da{} & & \da{} & & \da{} & & & & \\ 1 & \ra{} & I_K & \ra{} & G_K & \ra{} & \operatorname{Gal}(K^{ur}/K) & \ra{} & 1\\ \end{array}$$ We always have $\operatorname{Gal}(L^{ur}/L) = \operatorname{Gal}(K^{ur}/K) \simeq \hat{\mathbb{Z}}$. If $L/K$ is totally ramified, we have $I_L = I_K$ and I would love to conclude that $G_L = G_K$. But I am not sure whether this is possible or not.
Could you please help me resolving this problem? Thank you!
No. That $g\in G$, and $G/N \equiv H/N$ would imply that the coset $gN \subset H$, so that $g\in H$. This for all $g$.