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Hypothesis: Let $P$ be a finite $2$-group with two isomorphic normal subgroups $M$ and $N$ such that $P/M\cong C_4$ (the cyclic group of order $4$) while $P/N\cong C_2^2$. By the lattice theorem, there are exactly three groups $G$ with $N<G<P$.

Question: Is it possible that these three groups are all isomorphic?

Comments: There are plenty of examples satisfying the hypothesis. The smallest is $P=C_4\times C_2$, with $M=1\times C_2$ and $N=C_2\times 1$. In this case, of the three groups between $N$ and $P$, two are isomorphic to $C_4$ and one is isomorphic to $C_2^2$. Using magma, I have checked all groups $P$ of order at most $128$ and it seems that the answer to the question is "no" for these (barring a mistake in my code, which is entirely possible).

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  • $\begingroup$ I wonder if a 2-generated 2-group with all maximal subgroups isomorphic, and a non elementary abelian abelianized do the claim. Did you checked the free Burnside group on 2-generators and exponent 4. $\endgroup$ – Yassine Guerboussa Apr 16 '14 at 19:06
  • $\begingroup$ I'm not sure I understand your example. Could you maybe be more explicit? (Keep in mind that I am only considering finite groups.) $\endgroup$ – verret Apr 17 '14 at 1:55
  • $\begingroup$ This question has a somewhat similar flavor to mathoverflow.net/questions/153433/… $\endgroup$ – Russ Woodroofe Apr 17 '14 at 2:24
  • $\begingroup$ Ok, I think in a finite $p$-group $G$ such that $Aut(G)$ induces the full linear group on $G/\Phi(G)$, all the maximal subgroups of $G$ are isomorphic. Any relativley free finite $p$-group have the property above. I don't know if one can choose two maximal subgroups $N$ and $M$ such that the isomorphism between $N$ and $M$ (induced by an automorphism of $G$) takes a $G$ invariant maximal subgroup of $N$ whose quotient has exponent $4$ to a subgroup of $M$ whose quotient has exponent $2$. Another thing, any free Burnside group of exponent 4 is finite (thanks to Sanov). $\endgroup$ – Yassine Guerboussa Apr 17 '14 at 10:12
  • $\begingroup$ @YassineGuerboussa: If the isomorphism between maximal subgroups of $N$ and $M$ is induced by an automorphism of $G$ then the quotients will be isomorphic. Of course, this doesn't rule out the possibility that there's an isomorphism not induced by an automorphism of $G$, but according to my amateurish Magma calculations this isn't the case for the free Burnside group of exponent four on two generators. $\endgroup$ – Jeremy Rickard Apr 17 '14 at 12:46

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