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Let $G$ be a group and let $N$ be a normal subgroup of $G$. In addition suppose that $A$ is a subset of $G$ containing $N$, that is $N\subseteq A\subseteq G$. I wondering if it possible that $A/N$ form a group although $A$ doesn't form subgroup. My guess is that it is impossible, but i don't know how to prove it. Thanks!

p.s. In $A/N$ I mean $\{aN:a\in A\}$.

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  • $\begingroup$ You cannot prove it, because it is not true. $\endgroup$ – Dietrich Burde Oct 13 '16 at 12:28
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It is certainly possible, with the definition as you have written it -- consider $G = \mathbb Z/4\mathbb Z$, with $N = \{0,2\}$ and $A = \{0,1,2\}$. However, it is generally very unlikely, as there is no reason to assume why it would be so. In particular, for any $G$ you can take $N = \{1\}$, and in that case your question comes down to "is $A$ itself a subgroup"?

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Let $A'$ be a subgroup of $G/N$ and $\pi$ the projection $G \rightarrow G/N$. Chose any map $\sigma : G/N \rightarrow G$ such that $\pi \circ \sigma = id_N$ (as guaranteed by the axiom of choice). Note that it is easy to chose $\sigma$ such that $A = \sigma(A')$ satisfies your conditions.

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