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We have a group $G$ and $H$ is its normal subgroup. Say $K = \{C_1,...,C_k\}$ is a subgroup of $G/H$. Is the union C$_1 ∪···∪ C_k$ a subgroup of $G$?

I can see that $C_i$’s are elements of the factor group $G/H$ and thus $K$ is a union of cosets of $G$ wrt $H$. I also intuitively see that not any union of cosets is a subgroup, but I cannot deduce anything from $K$ being a subgroup though I think this fact is crucial

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The union of the cosets is the preimage of the subgroup $K$ of $G/H$ under the quotient map $G\to G/H$, and so it is a subgroup.

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  • $\begingroup$ Got there just a sec before me :) $\endgroup$ – Tim Raczkowski Apr 3 '16 at 19:25
  • $\begingroup$ @TimRaczkowski So assume I construct a map $G \to G/H$ such that if $C_i = g_iH$ then $g_i \to C_i$. But could you explain how exactly does it follow that union of elements of $C_i's$ form a subgroup of $G$? $\endgroup$ – tmac_balla Apr 3 '16 at 20:20
  • $\begingroup$ @tmac_balla Just use the map $g\mapsto gH$. $\endgroup$ – Tim Raczkowski Apr 3 '16 at 21:32
  • $\begingroup$ @TimRaczkowski yeah, that's what I am using. But by which theorem you deduce the final result? $\endgroup$ – tmac_balla Apr 3 '16 at 22:56
  • $\begingroup$ @tmac_balls If $f:G\to G'$ is a homomorphism, and $H$ a subgroup of $G'$, then $f^{-1}(G)$ is subgroup of $G$. You should be able to verify this by a routine check of the subgroup properties. $\endgroup$ – Tim Raczkowski Apr 3 '16 at 23:30

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