# Is a union of cosets of a normal subgroup a subgroup itself?

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

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.
• @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 $G$? – tmac_balla Apr 3 '16 at 20:20
• @tmac_balla Just use the map $g\mapsto gH$. – Tim Raczkowski Apr 3 '16 at 21:32
• @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. – Tim Raczkowski Apr 3 '16 at 23:30