Thread: internal semi-direct product with homomorphism

Let H be a cyclic group under addition. Let q=internal semidirect product between Zn and Z2. Let f: H to H be the inversion automorphism defined byf(x)=-x. Define a: Z2 to Aut(H) to be the homomorphism that maps a(1)=f. Show Zn qa Z2 is isomorphic to Dn by identifying two generators in Zn qa Z2 that satisfy the relations for Dn.
Note: (1) f is an element of Aut(H) and a is a well-defined homomorphism since f is an order 2 element of Aut(h).
(2) If H=Z, then Z qa Z2 is isomorphic to the infinite dihedral group.
I really have no idea how to find the generators in Zn qa Z2, anyhelp on this would be great!!!

Originally Posted by aabsdr

Let H be a cyclic group under addition. Let q=internal semidirect product between Zn and Z2. Let f: H to H be the inversion automorphism defined byf(x)=-x. Define a: Z2 to Aut(H) to be the homomorphism that maps a(1)=f. Show Zn qa Z2 is isomorphic to Dn by identifying two generators in Zn qa Z2 that satisfy the relations for Dn.
Note: (1) f is an element of Aut(H) and a is a well-defined homomorphism since f is an order 2 element of Aut(h).
(2) If H=Z, then Z qa Z2 is isomorphic to the infinite dihedral group.
I really have no idea how to find the generators in Zn qa Z2, anyhelp on this would be great!!!

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.

Tonio