I am wondering whether there is a counterexample which shows that subgroups and quotients don't determine the group.

More precisely, suppose there are two groups $G_1, G_2$ such that all of their proper non-trivial normal subgroups are 1-1 corresponding and if $1<H_1 < G_1, 1<H_2 < G_2$ are those proper normal subgroups that they correspond, then $H_1 \simeq H_2$, and $G_1 / H_1 \simeq G_2/H_2$. (Here $\simeq$ means isomorphic.)

Then $G_1 \simeq G_2$?

I guess it might be not true in general, but I don't know any nontrivial counterexample except the pair $(\mathbb{Z}_p, \mathbb{Z}_q)$.

Any comments on this will be highly appreciated!

  • $\begingroup$ Not to be annoying, but if $G_1/H_1 \simeq G_2/H_2$ for any proper $H_1, H_2$, then you can take $H_1 = H_2 = 0$ and conclude, so you should exclude trivial subgroups from the hypothesis. $\endgroup$ – guidoar Aug 8 '18 at 8:07
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    $\begingroup$ Since this seems to have been interpreted differently in the two answer, could you clarify if this is what you intended: Let $A$ and $B$ be the sets of proper non-trivial subgroups of the groups and assume we have a bijective map $f: A\to B$ such that we for all $H\in A$ have $H\cong f(H)$ and $G_1/H\cong G_2/f(H)$ whenever $H$ is normal? $\endgroup$ – Tobias Kildetoft Aug 8 '18 at 8:54
  • $\begingroup$ Following Tobias, what if the subgroups are not normal? Otherwise, why not simply take two simple groups? $\endgroup$ – verret Aug 8 '18 at 9:09
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    $\begingroup$ The question seems clear enough to me, but you need to specify that normal subgroups correspond to normal subgroup in the correspondence between subgroups, since otherwise you cannot form quotient groups. $\endgroup$ – Derek Holt Aug 8 '18 at 9:32
  • $\begingroup$ Thanks all. I corrected the original question as you commented. $\endgroup$ – user29422 Aug 8 '18 at 14:04

There is a counterexample of order $605 = 11^2 \times 5$ with the structure $11^2:5$. Let $$G = \langle x,y,z \mid x^{11}=y^{11}=z^5=1, xy=yx, x^z=x^4, y^z=y^5 \rangle,$$ $$H = \langle x,y,z \mid x^{11}=y^{11}=z^5=1, xy=yx, x^z=x^4, y^z=y^3 \rangle.$$ These are $\mathtt{SmallGroup}(605,5)$ and $\mathtt{SmallGroup}(606,6)$ in the small groups database.

They both have $121$ conjugate subgroups of order $5$, two normal subgroups of order $11$ with nonabelian quotients of order $55$, $10$ non-normal subgroups of order $11$, $22$ non-normal subgroups of order $55$, and one normal subgroup of order $121$.

  • $\begingroup$ Aha Derek! That is the answer by the real master here. Excellent. +1 from me. $\endgroup$ – Nicky Hekster Aug 8 '18 at 10:09
  • $\begingroup$ Wow! This is the answer what I wanted. Thank you very much!:) $\endgroup$ – user29422 Aug 8 '18 at 14:14

Smallest counterexample: $G_1=C_4$ generated by $x$ and $G_2=C_2 \times C_2$. Take $H_1=\langle x^2 \rangle$ and $H_2=C_2 \times \{1\}$.

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    $\begingroup$ The proper subgroups of these do not correspond one-to-one $\endgroup$ – Tobias Kildetoft Aug 8 '18 at 8:16
  • $\begingroup$ What does $C_n$ mean ?? $\endgroup$ – Anik Bhowmick Aug 8 '18 at 8:29
  • $\begingroup$ The cyclic group of order $n$ is denoted by $C_n$ (multiplicative notation) and this is also $\mathbb{Z}/n\mathbb{Z}$ (additive notation). But wait, Tobias triggered me: @user29422 what exaclty do you mean in your post by "such that all of their proper subgroups are 1-1 corresponding"? $\endgroup$ – Nicky Hekster Aug 8 '18 at 8:49
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    $\begingroup$ I added a comment on the question asking for a clarification from the OP. $\endgroup$ – Tobias Kildetoft Aug 8 '18 at 8:58

There are two (1) groups of order $4$. Each has a normal subgroup of order $2$ so the quotient groups are also order $2$. There is only one (1) group of order $2$ so the subgroups are isomorphic and the quotient groups are isomorphic.

(1) Up to isomorphism.

See The extension problem (Wikipedia)

  • $\begingroup$ The OP asked for the proper subgroups to correspond one-to-one $\endgroup$ – Tobias Kildetoft Aug 8 '18 at 8:16
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    $\begingroup$ @TobiasKildetoft Is there a technical meaning of "correspond" here that I am missing? $\endgroup$ – badjohn Aug 8 '18 at 8:21
  • $\begingroup$ Not sure what that would be. The two groups of order $4$ have different numbers of subgroups, so they don't correspond. $\endgroup$ – Tobias Kildetoft Aug 8 '18 at 8:22
  • $\begingroup$ @TobiasKildetoft I see what you mean now. So, it is a more interesting question than it first seemed. $\endgroup$ – badjohn Aug 8 '18 at 8:26

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