# Equal quotient groups

Suppose $G$ is a group, $H$ and $K$ are subgroups of $G$ and $N$ is a normal subgroup of $H$ and $K$. Suppose also that $H/N = K/N$. Does this mean that $H = K$?

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If you mean $H/N\cong K/N$, then no - consider the example of $$G=(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z}), \quad H=\{(0,0),(1,0)\},\quad K=\{(0,0),(0,1)\},\quad N=\{(0,0)\}$$ then $$H/N\cong H\cong \mathbb{Z}/2\mathbb{Z}\cong K\cong K/N,$$ but $H\neq K$.
If you really literally mean $H/N=K/N$, as an equality of underlying sets, then yes: the inverse image of $H/N=K/N$ under the quotient homomorphism $q:G\to G/N$ will necessary be equal to both $H$ and $K$, by the fourth isomorphism theorem, hence $H=K$.
Thanks for the reply. Yes, I meant that $H/N$ = $K/N$. –  Mikko Korhonen Oct 10 '11 at 15:40