Let's say we have a system of interacting particles that can divided into two populations. The symmetry group of each population is $G$, and the two populations are identical, so that I can exchange them and the system remains invariant (see figure). I think the symmetry under exchange of the two populations should be described by the symmetric group $S_2$, see here.

enter image description here

My question is: what is the symmetry group of the whole system? Intuitively it should be some kind of composition between $S_2$ and $G$, but which one?


If I understand your question correctly, you have two systems $A$ and $B$, on both of which a symmetry group $G$ acts, so we have mappings

$$G\times A\to A,\ \ \ (g,a)\mapsto ga$$


$$G\times B\to B,\ \ \ (g,b)\mapsto gb$$

If the composed system is $A\times B$, then it symmetry group obviously includes $G\times G$ by $(g,h)(a,b) = (ga, hb)$. We have an additional symmetry:

$$S_2\times (A\times B)\to A\times B,\ \ \ (\sigma, (a,b))\mapsto (b,a)$$

where $\sigma$ is the generator of $S_2$. Now what happens with the composition?

$$\sigma(g,h)(a,b) = (hb,ga)$$


$$(g,h)\sigma(a,t) = (gb,ha)$$

If we define the automorphism $\phi_\sigma$ of $G\times G$ by $\phi_\sigma(g,h) = (h,g)$, then we have

$$(g,h)\sigma(a,b) = (gb,ha) = \sigma\phi_\sigma(g,h)(a,b)$$

showing that your full symmetry group is (or rather contains) the semidirect product

$$(G\times G)\rtimes_\phi S_2$$

In the quantum mechanical context, $A$ and $B$ are vector spaces, and the combined system is something like $A\otimes B$, on which we also have the action of $G\times G$ and of $S_2$ and the final result is the same.

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  • $\begingroup$ Thank you very much doetoe! I'm not an expert of group theory, so please be patient, but I would like to ask you to comment why the symmetry group I'm looking for contains $(G\times G)\rtimes_\phi S_2$. Does it mean that $(G\times G)\rtimes_\phi S_2$ is not the definitive answer to my question? $\endgroup$ – user2983638 Jun 18 '15 at 15:44
  • $\begingroup$ @user2983638 I only wanted to say that the full symmetry group will contain this group, but it is not impossible that more new symmetries are introduced in going to the product of the systems. This is the best you can do if you don't know more about the systems. $\endgroup$ – doetoe Jun 18 '15 at 16:25
  • $\begingroup$ As a very simple example, assume that $A$ and $B$ each consist of 2 indentical independent subsystems $A = X_1\times X_2$, $B = X_3\times X_4$, and the symmetry group of each is $S_2$, interchanging these populations. Then the symmetry group of $A\times B$ indeed contains $(S_2\times S_2)\rtimes S_2$, generated by $(12), (34)$ and $(13)(24)$, but in fact it is $S_4$. $\endgroup$ – doetoe Jun 18 '15 at 16:25

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