In the second paragraph of the following paper, there is a statement: "Because the direct product of subgroups is automatically a subgroup.."


I don't see how that can be true...you can always take direct product of a subgroup with itself many times and create a group of order larger than the parent group...

  • $\begingroup$ My understanding is that the definition excludes taking direct products of subgroups with themselves. Generally subgroups of a group are distinct. $\endgroup$ – supercooldave Jun 24 '10 at 7:21
  • $\begingroup$ unless it's the trivial subgroup ;) $\endgroup$ – Pietro Majer Jun 24 '10 at 7:55
  • $\begingroup$ @SCDave: but you could take a product of a subgroup with a subgroup of that subgroup. Or take a product of two subgroups that don't commute. The sentence in question (I've just looked) is a little odd as it reads as though it is a general statement that always applies, which can't be true. However, if it is to be read as only applying to the situation of the paragraph then it makes a little more sense because the subgroups in question have coprime order. (I'm not saying that that settles the matter, but that's my first impression.) $\endgroup$ – Loop Space Jun 24 '10 at 8:03
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    $\begingroup$ I agree that the statement is incorrect or meaningless taken at face value. My best guess is that the "direct product" of two subgroups $H_1$ and $H_2$ of a group $G$ is the set $H_1 H_2$ under the conditions: $H_1 \cap H_2 = \{e\}$ and every element of $H_1$ commutes with every element of $H_2$. At least it is true that in this case $H_1 H_2$ is a subgroup which is canonically isomorphic to $H_1 \times H_2$. $\endgroup$ – Pete L. Clark Jun 24 '10 at 9:11
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    $\begingroup$ @Pete: I disagree! This was the question that the person was stuck on in trying to figure out something (which may well have been the finite subgroups of U(3)) and so this was the right question to ask. Working through a paper and trying to really understand everything is a very good skill to learn. The question "What are the finite subgroups of U(3)?" would mean that they probably wouldn't follow-up by working through this paper carefully. $\endgroup$ – Loop Space Jun 24 '10 at 11:15

The paper's stated goal is to check that certain proposed ``subgroups'' are not actually subgroups at all. The proposed subgroups are groups that are direct products A × B. To show that A × B is not a subgroup, it suffices to show that A × 1 is not a subgroup, and that is the strategy taken in the paper and described by that sentence. They show that A is not isomorphic to a subgroup of SU(3), and so A × B is not isomorphic to a subgroup of SU(3).

I believe the statement they are making in symbols is: A ≤ A×B. In other words, they have stated the converse of the statement they use.

They also probably intend for B to be embedded in the center of SU(3), eliminating any worries about the direct product. Here they may also be using ⊗ to denote a mixture of the direct product and the Kronecker product. Even if A ∩ B ≠ 1, A ⊗ B = { ab : a in A, b in B } when B is central in SU(3).

It may be wise to avoid taking a sentence from a physics paper out of context and deriving logical consequences from it. This sentence is likely meant to explain a strategy, not state a mathematical truth. You should examine the sentence, but don't worry if you can derive contradictions from assuming the sentence is true (which would cause great worry in a math paper).

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    $\begingroup$ +1 particularly for the last paragraph! $\endgroup$ – Loop Space Jun 24 '10 at 13:28
  • $\begingroup$ @Jack First,physicists are generally notorious for not having total logical consistency in thier papers about mathematical topics. Secondly,taking direct products of groups,while the simplest and one of the most useful ways of generating new groups,does have interesting perils that aren't immediately obvious and this is a good example of this. $\endgroup$ – The Mathemagician Jun 24 '10 at 19:26

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