> Looking at the code, I assume what makes the difference is that without 
> xsl:for-each-group, all "b" and "c" elements within their parent element 
> would go into the same container1 instance instead of each "b"/"c" 
> sequence receiving their own container. Am I right?

the usage of for-each-group that you suggested, with a constant grouping
key, is exactl_ the same as the version I posted that did not use
for-each-group at all. In both cases they will only make a single
container1 element and place into that all of the relevant elements even
if they occur later in the input sequence.

the "two methods" that i posted in the earlier reply were designed to
handle this case, they will separately group each contiguous run of b,c
and d,e,f groups into possibly multiple container element_s_. 


> If so, it won't hurt me, as my "b" and "c" elements will always appear 
> before any "d" element, therefore I think can do without
> xsl:for-each-group.

so, you don't need for-each groop here.

> The lesson I learned: The hardest part with XSLT (2.0) is to find its 
> easy solution to your complex problem. ;-)

sounds about right!

David