This is a very confusing (confused?) question:
(i) It seems to be that ;
(ii) Internal (semi)direct product exists when we form it from two subgroups identified in some given group. Which group are taken from so that we can
talk of their internal semidirect product??
(iii) We have the semidirect product , where acts on by means of the inversion automorphism: ( against my
instincts, I write instead of since the OP insists, apparently, on an additive notation).
From the above we see that H is...inexistent, or if you prefer unnecessary.
So let's say we have the outer semidirect product , where .
Suppose , with both written additively modulo n, 2, resp.
Define . It's easy to see that ( multiplicatively one'd write this as ) , and
, and from here we have that .
As you can see, insisting in an additive notation can make things messier since we "sum" elements but his sum isn't, of course, commutative...
I'd rather had taken , multiplicative cyclic groups of ordern, 2 resp., and do all the above we usual multiplicative notation.
Of course, if instead the cyclic group of order n you take the infinite cyclic group then the above shows that we get the infinite dihedral group......and after reading the last part of your post I now understand what you surely meant with that H at the beginning, but STILL the question is presented in a rather sloppy way.