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Math Help - internal semi-direct product with homomorphism

  1. #1
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    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!!!
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  2. #2
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    Quote Originally Posted by aabsdr View Post
    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
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