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If $G$ is a finite group such that their is a normal subgroup $H$ of $G$ such that $G/H \cong V_4$ , where $V_4$ is the Klein four group , then is it true that $G$ can be written as a union of three proper subgroups ?

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Yes: let $K,L,M$ the three proper subgroups of $V_4$, then $G=\pi^{-1}(K)\cup\pi^{-1}(L)\cup\pi^{-1}(M)$ (where $\pi$ is the quotient projection $x\mapsto x+H$) because $V_4=K\cup L\cup M$ and the counterimage of a proper subgroup under a surjective homomorphism is proper.

(In fact $\pi(\pi^{-1}(K))=K\cap\pi(G)=K$)

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  • $\begingroup$ Pullback of a proper subgroup is proper subgroup ? $\endgroup$
    – user228168
    Commented Apr 20, 2015 at 14:57
  • $\begingroup$ @SaunDev I was a little imprecise. The counterimage of a proper subgroup is proper if the homomorphism is surjective (which is the case), because in general $f(f^{-1}(L))=L\cap f(G)$. I'll edit $\endgroup$
    – user228113
    Commented Apr 20, 2015 at 15:05
  • $\begingroup$ Do we need $f(f^{-1}(L))$ or $f^{-1}(f(L))$ ? $\endgroup$
    – user228168
    Commented Apr 20, 2015 at 15:08
  • $\begingroup$ $f(f^{-1}(L))$. If by chance $f^{-1}(L)=G$, then $f(f^{-1}(G))=f(G)=$ the whole image, which in this case is $V_4$. But we know that for a surjective homomorphism $f(f^{-1}(L))=L$, which yield a contraddiction with the fact that $L$ was proper to begin with. @SaunDev $\endgroup$
    – user228113
    Commented Apr 20, 2015 at 15:10

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