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If |G/H|=4 and G/H is not cyclic then G is union of three proper subgroups. I understand that there are three subgroups of order two in the G/H group. Consequently, in the group G there are three subgroups A, B, C of index two which in them pass under the homomorphism. But how to prove that G is a union of the subgroups A, B, and C

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    $\begingroup$ If $G/H$ is cyclic, it has only one subgroup of order $2$... $\endgroup$ Commented Apr 9, 2018 at 16:51
  • $\begingroup$ yes, but I meant that G/H is not cyclic $\endgroup$
    – Cat
    Commented Apr 9, 2018 at 16:59

1 Answer 1

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Hint: If $A,B,C$ are subsets in $G$, then $$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|.$$

$G/H\cong K_4$, so you know that by the Correspondence theorem that there exists subgroups $A,B,C$ in $G$ such that $|A|=|B|=|C|=2|H|$ and $|A\cap B|=|B\cap C|=|C\cap A|=|A\cap B\cap C|=|H|$, so what can you say about $|A\cup B\cup C|$?

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  • $\begingroup$ Why do you think that |A∩B|=|B∩C|=|C∩A|=|A∩B∩C|=|H| ? We can say that A∩B is a subgrop A and B . $\endgroup$
    – Cat
    Commented Apr 9, 2018 at 17:54
  • $\begingroup$ @Cat you can pick three proper subgroups in the quotient groups whose union is the entire Q/H. The quotient is the Klein 4 group. You pick three subgroups in Q/H such that each two intersect trivially. They correspond to three subgroups of Q whose union is entire Q. You prove this by correspondence theorem and by counting the elements in each subgroup. $\endgroup$
    – Frenzy Li
    Commented Apr 9, 2018 at 18:00
  • $\begingroup$ The point is that the intersection of each two of these three proper subgroups in G is H, just like the intersection of each two proper subgroups of G/H is the identity. $\endgroup$
    – Frenzy Li
    Commented Apr 9, 2018 at 18:12
  • $\begingroup$ let V4=K∪L∪M and π−1(K)=A, π−1(L)= B andπ−1(M)=C. Let π(G) = V4 but that does not mean that G=π−1(K)∪π−1(L)∪π−1(M) $\endgroup$
    – Cat
    Commented Apr 9, 2018 at 18:17
  • $\begingroup$ By the correspondence theorem, the three different proper subgroups of $K_4$ correspond to three different proper subgroups $K,L,M$ of $G$ that contain $H$. Each of $K,L,M$ are different and are of index 2 in $G$. The intersection of each of $K,L,M$ must be a proper subgroup of $G$. The intersections must be index 2 in $K,L$ or $M$ (K=L otherwise, for example, which leads to contradiction). $\endgroup$
    – Frenzy Li
    Commented Apr 9, 2018 at 18:24

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