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I remember at some point being told that there is a way to tell if a subgroup is normal by looking at the lattice of subgroups of the whole group, but I can't seem to remember how this goes. Could someone remind me please?

user140776
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    Unless there is a special way to denote them in the lattice, and that'd be more a characteristic of whoever writes down the lattice, there is no way to tell them apart in the general case – DonAntonio Oct 15 '16 at 21:35
  • As a consequence of the Sylow theorems, there can exist at most 1 normal subgroup of any size $p$, where $p$ is prime, since otherwise two normal subgroup would be conjugate (a contradiction). This doesn't cover all normal subgroups (non-Sylow normal) but I think the concept applies generally. – Phillip Hamilton Oct 15 '16 at 21:40
  • can one identify the conjugacy classes from looking at the lattice? – user140776 Oct 15 '16 at 21:42
  • What I was thinking specifically was that if you look at the lattice and there is only one subgroup of some order $p$, I think by Sylow you would immediately know it is normal – Phillip Hamilton Oct 15 '16 at 21:44
  • But you can't tell the order of a subgroup by looking at the lattice. – Derek Holt Oct 15 '16 at 21:51
  • I suppose we can assume that the nodes of the lattice are labeled with the orders of the subgroups they represent. – user140776 Oct 15 '16 at 21:53
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    @PhillipHamilton The group $;\Bbb Z_p\times\Bbb Z_p\times\ldots\Bbb Z_p;$ has lots of normal subgroups of order $;p;$ ... and you can change there one (or more but...generalize) of the factors and still get several normal subgroups of order $;p;$ . – DonAntonio Oct 15 '16 at 22:16
  • @DonAntonio oh duh...I forgot that Sylow-p subgroups may not have order $p$, they have order $p^n$ generally. So it would be even harder to check from the lattice. However, if you see just one subgroup of order $p$, it should still follow it's normal, would it not? This check is not exhaustive and doesn't identify all normal subgroups, but it will identify some. The OP question isn't necessarily looking for completeness. – Phillip Hamilton Oct 15 '16 at 22:24
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    The relevance of the comments (including the OP's comments) to the question are becoming increasingly unclear. Can someone please give an example of two groups $G$ and $H$ and a lattice-isomorphism between the subgroups of $G$ and $H$ that does not put normal subgroups of $G$ in 1-1 correspondence with normal subgroups of $H$. (i.e., I am asking for a proof of the claim in Don Antonio's first comment). If such an example does not exist, then lattice properties don't distinguish the normal subgroups and the idea of lattices decorated with the sizes of the subgroups is worth discussing. – Rob Arthan Oct 15 '16 at 22:32
  • If $p$ is a prime and $q$ is a prime dividing $p-1$, then the elementary abelian group of order $p^2$ and the nonabelian group of order $pq$ both have the same subgroup lattice: the full group, the trivial group, and $p+1$ other groups, with no inclusion except the trivial ones. (For a specific example, consider $Z_3^2$ and $S_3$.)

    In the first one, all subgroups are normal, but not in the second one.

     – verret Oct 18 '16 at 22:07
  • In particular, the answer to the question as posed is "no". I'm not sure about the answer if we decorate the subgroups with their order. I'll try to think about this. – verret Oct 18 '16 at 22:25
  • I think you might have heard the fact the GAP tools to display the subgroup lattice display normal subgroups differently and group conjugacy classes together - see e.g. http://math.stackexchange.com/q/1737962/ and http://math.stackexchange.com/q/1561370/ – Olexandr Konovalov Oct 19 '16 at 20:19

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This is not possible.

If $p$ is a prime and $q$ is a prime dividing $p−1$, then the elementary abelian group of order $p^2$ and the nonabelian group of order $pq$ both have the same subgroup lattice: the full group, the trivial group, and $p+1$ other groups, with no inclusion except the trivial ones. (For a specific example, consider $Z_3^2$ and $S_3$.) In the first one, all subgroups are normal, but not in the second one.

Even decorating the subgroups with their order does not help:

The groups with GAP ID [16,5] and [16,6] both have 11 subgroups and, in fact, have isomorphic subgroup lattices. (I checked this with a mix of computer and hand calculations, so caveat emptor.) Since they are $p$-groups, the order of every subgroup can be recovered just from the lattice.

On the other hand, [16,5] is abelian (it is isomorphic to $Z_8\times Z_2$) and has every subgroup normal, but [16,6] has non-normal subroups (it is sometimes called $M_{16}$, see http://groupprops.subwiki.org/wiki/Groups_of_order_16).

I suppose the next question would be: what if we are also given the isomorphism type of the full group? In other words, can one find an example of a group such that its subgroup lattice, decorated with the subgroup orders, admits an automorphism mapping a normal subgroup to a non-normal one?

My guess is that the answer is yes...

verret
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