How Do Elements of Direct Product Groups Commute in Particle Physics?

[foobar]

From Gauge Theory of particle physics, Cheng and Li I don't understand the flollowing:

"Given any two groups G={g1,..} H= {h1,h2,...}
if the g's commute with the h's we can define a direct product group G x H={$$g_ih_j$$} with the multplication law:
$$g_kh_l . g_mh_n = g_kh_m . h_lh_n$$

Examples of direct product groups are SU(2) x U(1) ( the group consists of elements which are direct products of SU(2) matrices and the U(1) phase factor) and
SU(3) x SU(3) (the group consists of elements which are direct products of matrices of two different SU(3)'s."

My question is: You need the g's and h's to commute. I can see how SU(2) and U(1) elements can commute. I don't see how an element of SU(3) commutes with
another SU(3) element?

 

[Hurkyl]

"Given any two groups G={g1,..} H= {h1,h2,...}
if the g's commute with the h's

What he means is if they commute in the group we're about to form.

This statement, as written, doesn't make sense: if all we have is a group G and a group H, then we don't have a way to multiply an element of G with an element of H. But in the process of building a new group, we have lots of freedom to choose how such a multiplication might make sense.

I don't see how an element of SU(3) commutes with
another SU(3) element?

In particular, the issue here is that the elements of the different 'copies' of SU(3) commute, simply by the definition of the product group. If I paint one copy red and another copy blue, then we can have both
$${\color{red} x y} \neq {\color{red} y x}$$
and
$${\color{red} x} {\color{blue} y} = {\color{blue} y} {\color{red} x}$$
because the product is defined differently for elements of the same color and of opposite colors.


An isomorphic way to define GxH is as the group whose elements are ordered pairs (g, h) where g is in G and h is in H, and whose multiplication is performed pointwise.

i.e. (g,h)(g',h') = (gg',hh')

We can then identify each element g of G with its image (g,1) in GxH. Similarly, we can identify h with (1,h). If we do this, note that the g's commute with the h's under this convention:

(g, 1) (1, h) = (1, h) (g, 1)

 

[foobar]

thanks for that.

But still was a bit confused.But found this in a set of lecture notes on group theory( which I quit):

elements of A x B were written (a,b)
can write this as (a,b) = $(a,e_B)(e_A,b)=(e_A,b)(a,e_B)=(ae,eb)=(a,b)$

shorthand notation for elements of A x B
ab=ba letting [tex](a,e_B) \rightarrow "a" [/itex]<br /> $and (e_A,b) \rightarrow "b"$[/tex][itex][/itex]