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Let $G$ be a finite simple group in which there is no element of order $p^2$ for all primes $p\mid\vert G\vert$. Suppose that $H$ is a finite group whose number of nontrivial proper subgroups is as same as $G$ and there is a bijection $\phi$ from the set of nontrivial subgroups of $G$ to the set of nontrivial subgroups of $H$ such that for every two subgroups of $G$ say $K_{1}, K_{2}$ we have $K_{1}\leq K_{2}$ if and only if $\phi(K_{1})\leq \phi(K_{2})$(i.e. $\phi$ preserves inclusions)

Can we say that $G\cong H$?

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  • $\begingroup$ Are there even such simple groups? Barring the abelian ones I mean? $\endgroup$ – Tobias Kildetoft Jan 2 '17 at 8:51
  • $\begingroup$ @TobiasKildetoft: Perhaps, $A_5$ fits? $\endgroup$ – Ilya Bogdanov Jan 2 '17 at 9:16
  • $\begingroup$ @IlyaBogdanov Ahh, right. So apart from that one? My guess would be that as soon as they get large enough, there will be at least one prime for which this fails (but I didn't actually check). $\endgroup$ – Tobias Kildetoft Jan 2 '17 at 9:19
  • $\begingroup$ @TobiasKildetoft: I'm not so sure. In partivular, $|PSL(2,\mathbb F_p)|$ may have a few nontrivial prime power factors... $\endgroup$ – Ilya Bogdanov Jan 2 '17 at 9:22
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    $\begingroup$ Regarding which simple groups are possible: The Sylow $2$-subgroups of $G$ are of exponent 2, and so are abelian. Thus Walter's classification applies, and we know that $G$ is either ${\rm PSL}_2(q)$ (with $q$ even or equivalent to $3,5\pmod 8$) or is $J_1$ or is ${^2G_2(q)}$. I think the linear groups give examples (depending on the prime decomposition of $q+1$ and $q-1$), the ATLAS tells us that $J_1$ is also an example, and finally I think ${^2G_2}(q)$ usually has elements of order $9$, but I'd need to double check.... $\endgroup$ – Nick Gill Jan 2 '17 at 10:59
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The answer is "yes". Indeed, much more is true. Refer to the following paper:

Michio Suzuki, MR 39717 On the lattice of subgroups of finite groups, Trans. Amer. Math. Soc. 70 (1951), 345--371.

The bijection that is mentioned by the OP is a lattice isomorphism. In the cited paper, Suzuki proves that if $G$ is simple and is lattice isomorphic to a group $H$, then $H$ is simple. This result does not use the Classification of Finite Simple Groups.

Now, using the Classification of Finite Simple Groups, one can conclude that $G\cong H$. This last statement can be found on p.434 in:

Groups and lattices, Palfy, P. In the Proceedings of Groups St Andrews 2001.

None of these results require the restriction on element orders given in the first sentence of the question.

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