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The way I think of the semidirect product is as a "deformation" of the direct product. Is there a way of making this intuition precise? Perhaps using some certain (co-) homology theory of groups?

A B C
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  • What's wrong with the simple fact that the direct product is a special case of the semidirect product? – Zhen Lin Jul 30 '14 at 21:33
  • I think that s/he would agree with this statement, but they still want an answer to a sort of converse to your statement: Is any semidirect product a deformation of a direct product? (where obviously any direct product is trivially (by definition) a special sort of semidirect product). – Squirtle Jul 30 '14 at 21:38
  • @ZhenLin: well if this were not the case, it would not make much sense, would it? There is nothing wrong with that, I am just curious whether there is some way of measuring (in terms of some group) the difference between them. I am well aware that the semidirect product generalizes the direct product, and facts such as that the semidirect product of $G$ and $H$ need not be abelian even if $G$ and $H$ are. To me this naturally suggests to think of the semidirect product as a kind of (possibly noncommutative deformation) of the direct product, in analogy to QM as a deformation of cl. mechanics. – A B C Jul 30 '14 at 21:41
  • Well perhaps this analogy is not a very good one, but I think there are certain kinds of similarities. Perhaps it is not a very important question in the theory of groups, as I am certain that one would otherwise set up some kind of homological apparatus – A B C Jul 30 '14 at 21:43
  • I think this is the wrong way to think about the semidirect product. In fact the semidirect product is not a product at all. See http://math.stackexchange.com/questions/867203/describing-the-wreath-product-categorically/867292#867292. – Qiaochu Yuan Jul 30 '14 at 21:59
  • @QiaochuYuan: I am (of course) not claiming that the semidirect product is a product in the categorical sense, and it is not clear to me why this should contradict the deformation intuition, after all the it would be odd to call (as I do) something a deformation if it were to retain all properties (and that the direct product is the categorical product is its defining property, so its not just "some" property) of the things deformed. I think considering short exact sequences would rather support the deformation viewpoint. – A B C Jul 30 '14 at 22:09
  • Well, to be more precise: the semidirect product $G \rtimes_\phi H$ is the product $G \times H$ "deformed" by the parameter $\phi : H \to \mathrm{Aut}(G)$, and if you set $\phi$ to be the constant function with value $\mathrm{id}$, then you get back the ordinary direct product. How is this not already analogous to deformation quantisation? – Zhen Lin Jul 30 '14 at 22:30
  • @ZhenLin: I totally agree. So in a sense what I am asking for is: what is $x$ in Deformation quanziation/Hochschild cohomology = Direct product deformation/$x$. (Well "direct product deformation" is perhaps a silly name, but w/e.) – A B C Jul 30 '14 at 22:56

2 Answers2

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This was supposed to be a comment but it got too long.

It is misleading to think of the semidirect product with $\varphi = \text{id}$ as the categorical product. As I explain in this answer, the semidirect product is a (higher) colimit rather than a limit, and in particular its (higher) universal property involves maps out of it rather than maps into it. One way to state this universal property when $\varphi = \text{id}$ is that $G \times H$ is the commutative coproduct of the groups $G$ and $H$: that is, it is the universal group equipped with maps from $G$ to $H$ whose images commute.

This is more visible once we generalize the semidirect product to other contexts. For example, if $G$ is a group acting on a $k$-algebra $R$, then there is a notion of crossed product algebra

$$R \rtimes_{\varphi} G \cong R \langle G \rangle / (grg^{-1} = \varphi(g) r)$$

which is a version of the semidirect product. Here by $R \langle G \rangle$ I mean a "noncommutative group algebra" constructed by starting from $R$ and adjoining noncommuting formal variables, one for each element of $G$, subject to the relations in $G$. When $\varphi = \text{id}$ this reproduces the tensor product $R \otimes_k k[G]$, which again is the commutative coproduct of algebras (and in particular is neither the product nor the coproduct).


Having said that let me try to answer the original question. From the deformation perspective, there's no reason to restrict our attention to semidirect products: more generally we can consider arbitrary short exact sequences

$$1 \to N \to X \to G \to 1$$

as deformations of the trivial short exact sequence

$$1 \to N \to N \times H \to H \to 1$$

and ask if we can come up with a cohomology theory that describes this. The answer is yes. The thing we want to compute, which classifies the possible choices of $X$, is the nonabelian cohomology

$$H^1(G, \underline{\text{Aut}}(N)) \cong [BG, B \underline{\text{Aut}}(N)]$$

of $G$ with coefficients in the automorphism 2-group of $N$ (here $B$ denotes taking the classifying space). This is a gadget which tracks not only the automorphisms of $N$ but also the auto-natural transformations among those automorphisms, thinking of $N$ as a one-object category and automorphisms of $N$ as functors from that category to itself. It can be thought of as a topological group with $\pi_0$ isomorphic to $\text{Out}(N)$ and $\pi_1$ isomorphic to $Z(N)$.

The standard story in group cohomology makes extra simplifying assumptions to avoid the terror of having to talk about 2-groups. The easiest extra assumption is to assume that $N$ is central (and in particular abelian). This turns out to free us from having to think about $\text{Out}(N)$, and now the resulting central extensions are classified by the ordinary group cohomology

$$[BG, B^2 N] \cong H^2(G, N).$$

This whole story is better thought of as an aspect of homotopy theory. Short exact sequences of groups correspond to fiber sequences

$$BN \to BX \to BG$$

of homotopy types, and in general the classification of fiber sequences with fiber (homotopy equivalent to) $F$ (here $BN$) and base $B$ (here $BG$) is given by the nonabelian cohomology

$$[B, B \text{Aut}(F)]$$

of $B$ with coefficients in the topological group of self-homotopy equivalences of $F$. See also principal bundle.

Qiaochu Yuan
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Let $H,K$ be subgroup of $G$ such that $G=HK$ and $H\cap K=e$.

case 1: $H,K$ are normal in $G$.

In that case as we can easily show that $hk=kh$ for $h,k\in H,K$. That is why giving meaning $hk$ to $(h,k)$ is very natural.

case 2: Now it is time ask what if only $H$ is normal? Now we can not say $hk=kh$ but we can say that $H^k=H$ or we have an homomorphism $\phi: K\to Aut(H)$ and surprisingly $\phi$ completletly determines the result of "$hk$" and vice versa and $K$ is also normal if and only if $\phi$ is trivial homomorphism.

case 3: Now we removed all normality case which is known as Zappa–Szép product which define general method to define "$hk$".

mesel
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  • Well this is obviously not what I am asking for. – A B C Jul 30 '14 at 22:57
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    you can not read this passage in $11$ sec. – mesel Jul 30 '14 at 22:58
  • By the way the downvote is not mine (I cannot even downvote). But what you are pointing out is already known to me, and can be found in every undergrad algebra textbook. I am asking for something slightly more advanced. – A B C Jul 30 '14 at 23:01
  • I am not interested which books write this fact, you asked the intuation ,I was trying to write the intuation. – mesel Jul 30 '14 at 23:05