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Let $G$ be a finite group and $M$ be a nontrivial proper subgroup of $G$ with the following conditions:

a) If $H$ is a subgroup of $G$ such that $M\lneqq H\lneqq G$, then $H$ contains at least one minimal subgroup of $G$ say $L$, such that $M\cap L=1$.

b) If $K$ is a subgroup of $G$ such that $M\cap K=1$ then $K\cong Z_p$ for some prime $p$.

Can we say that $M$ is a maximal subgroup of $G$?

Please read my comments after Derek Holt's answer.

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    $\begingroup$ Counterexample: $p$ odd, $G$ the nonabelian group of order $p^3$ and exponent $p$, and $M$ its center (of order $p$). Then (a) is trivial (it clerly holds in any group of prime exponent) and (b) is clear since every subgroup of $G$ with order $p^2$ contains the center. $\endgroup$
    – YCor
    Commented Oct 9, 2016 at 3:57
  • $\begingroup$ I think if we add the following condition to the question, then we probably exclude your Counterexample and other similar cases. c) There exists at least one maximal subgroup say $M^{'}$, such that $M\nleq M^{'}$ and $M^{'}$ contains all minimal subgroups of $M$. $\endgroup$
    – H.Shahsavari
    Commented Oct 9, 2016 at 7:52
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    $\begingroup$ Could you edit your question to make it clear? modifying the question in the comments is confusing. Also opening a bounty for a question ticked as answered is somewhat contradictory. $\endgroup$
    – YCor
    Commented Oct 17, 2016 at 7:02
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    $\begingroup$ The current answer is perfectly good and only becomes "out-of-date" if you change the question. Which you really shouldn't (especially in comments to the accepted answer), except for adding condition c) (and saying so in a comment) as YCor already asked. $\endgroup$
    – Francois Ziegler
    Commented Oct 17, 2016 at 7:43
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    $\begingroup$ Constantly moving the goalposts is generally frowned upon across all SE sites. It's okay to make separate, new questions for each new set of goalposts, though since this is a research-level site you might want to consider the possibility that you do not currently understand whatever motivates your question well enough to know what the correct question to ask is. $\endgroup$
    – zibadawa timmy
    Commented Oct 17, 2016 at 9:10

1 Answer 1

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I remember answering a very similar (possibly the same) question recently but I cannot find it.

Let $G$ be a central product of a cyclic group $M$ of order $4$ and the dihedral group $D_8$ of order $8$. So $|G|=16$. (In fact you get an isomorphic group if you replace $D_8$ by $Q_8$. A central product of $C_4$ with an extraspecial $2$-group is called a group of symplectic type.)

Then all subgroups of $G$ strictly between $M$ and $G$ have order $8$ and are isomorphic to $C_4 \times C_2$, and have the $C_2$ as the required subgroup $L$. But $M$ has no complement in $G$, so any subgroup $K$ with $K \cap M=1$ has order $1$ or $2$. (You need to assume that $K$ is nontrivial in the problem.) This example also satisfies your condition c), since the subgroups $D_8$ (and $Q_8$) are maximal and contain all minimal subgroups of $M$.

You can construct similar examples for any prime $p$ as a central product of $C_{p^2}$ with an extraspecial group of order $p^3$.

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  • $\begingroup$ Dear all. After a long check I got some new results which may leads to get a corrected version of this question. Now I invite you all to think of this new version and if possible help me to find a useful result. Let $G$ be a finite group and $L_{1}, L_{2},...,L_{n}$ be a family of minimal subgroups of $G$. Suppose that $M_{1},..., M_{k}$ are the all proper subgroups of $G$ which contain $L_{i}$'s as their minimal subgroups and have not any minimal subgroups other than $L_{i}$'s. $\endgroup$
    – H.Shahsavari
    Commented Oct 17, 2016 at 6:36
  • $\begingroup$ Also suppose that the following conditions hold for every $M_{j}$, $1\leq j\leq k$: a) If $H$ is a subgroup of $G$ such that $M_{j}\lneqq H\lneqq G$, then $H$ contains at least one minimal subgroup of $G$ say $L^{'}$, such that $M_{j}\cap L^{'}=1$. b) If $K$ is a subgroup of $G$ such that $M_{j}\cap K=1$ then $K\cong Z_p$ for some prime $p$. Can we say that at least one of the $M_{j}$'s is a maximal subgroup of $G$? $\endgroup$
    – H.Shahsavari
    Commented Oct 17, 2016 at 6:39
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    $\begingroup$ @H.Shahsavari: comments are not meant to modify questions. It is up to you to make your question clear in the question box. In the present case with an accepted answer, it would probably be best to ask another question, with due reference to this one. $\endgroup$
    – Benoît Kloeckner
    Commented Oct 17, 2016 at 7:31
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    $\begingroup$ @BenoîtKloeckner It is starting to look as though this process will go indefinitely: counterexample leads to extra hypotheses. $\endgroup$
    – Derek Holt
    Commented Oct 17, 2016 at 8:04
  • $\begingroup$ َAs it seems, no one want to answer the question and every body want just to blame me. OK keep on nagging. $\endgroup$
    – H.Shahsavari
    Commented Oct 17, 2016 at 17:51

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