# Do coset representatives of a normal group form a subgroup of G?

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?

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.
• @DustanLevenstein One sufficient condition is that $\langle K \rangle \cap N=\{1\}$. – Vipul Kakkar Dec 6 '13 at 17:38