# Math Help - internal semi-direct product with homomorphism

1. ## 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!!!

2. 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 $H=\mathbb{Z}_n$ ;

(ii) Internal (semi)direct product exists when we form it from two subgroups identified in some given group. Which group are $\mathbb{Z}_n\,,\,\mathbb{Z}_2$ taken from so that we can

talk of their internal semidirect product??

(iii) We have the semidirect product $\mathbb{Z}_n \rtimes_a \mathbb{Z}_2$ , where $a$ acts on $\mathbb{Z}_n$ by means of the inversion automorphism: $a\cdot k:= -k\,,\,k\in\mathbb{Z}_n$ ( against my

instincts, I write $a\cdot k$ instead of $k^a$ 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 $K:=\mathbb{Z}_n \rtimes_a\mathbb{Z}_2$ , where $a\cdot k:= -k\,,\,\,\forall k\in\mathbb{Z}_n$.

Suppose $\mathbb{Z}_n=<1_n=1\!\!\!\pmod n>=\{0,1,2,\ldots,n-1\}\!\!\!\pmod n\,,\,\,\mathbb{Z}_2$ $=<1_2=1\!\!\!\pmod 2>=\{0,1\}\!\!\!\pmod 2$ , with both written additively modulo n, 2, resp.

Define $s:=(0,1_2)\,,\,t:=(1_n,0)\in K$ . It's easy to see that $2s=nt=(0,0)$ ( multiplicatively one'd write this as $s^2=t^n=1$ ) , and

$sts=(0,1_2)+(1_n,0)+(0,1_2)=(-1_n(=n-1),1_2)+(0,1_2)=(n-1,0)=-t$ , and from here we have that $K\cong D_{2n}$ .

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 $C_n\rtimes C_2$ , 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