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Let G be a FINITE group and N is a normal subgroup of G . Let K be a set of representatives of the cosets of N . Is K a subgroup of G ? is there any specific condition for them to form a subgroup of G?

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Not in general, no. The existence of a choice of representatives that form a subgroup is exactly the condition for $G$ to be a semidirect product of $N$ and $G/N$.

The answer is still the same for finite groups. The smallest counterexample is the cyclic group of order 4, $G = \langle \sigma \rangle$ with $\sigma$ order 4, and $N = \langle \sigma^2\rangle$. Then $G/N$ is the cyclic group of order 2, but the coset representatives of the nontrivial element of $G/N$ are $\sigma$ and $\sigma^{-1}$, neither of which has order 2.

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  • $\begingroup$ please give me a reference with a proof for this $\endgroup$ – user20216 Dec 6 '13 at 16:39
  • $\begingroup$ Have you seen semidirect products before? It might be a good exercise for you to figure out a proof of this yourself. $\endgroup$ – Dustan Levenstein Dec 6 '13 at 16:41
  • $\begingroup$ I just wanted you to recommend a book for this topic $\endgroup$ – user20216 Dec 6 '13 at 16:43
  • $\begingroup$ Artin is a good introductory algebra text if that's what you're asking for. I don't have a copy handy to check whether it has this proposition specifically, but it shouldn't be too far off; this characterization of semidirect product is essentially the motivation that is typically given for the definition in the first place. $\endgroup$ – Dustan Levenstein Dec 6 '13 at 16:46
  • $\begingroup$ @DustanLevenstein One sufficient condition is that $\langle K \rangle \cap N=\{1\}$. $\endgroup$ – Vipul Kakkar Dec 6 '13 at 17:38

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