Weird Semidirect Product Formula

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  • #1
jstrunk
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TL;DR Summary
I can't understand how to use the weird formula in my book
My book gives this formula for the semidirect product for groups ##Z_p## and ## Z_q## for primes p<q and p divides (q-1).
##(a,b)*(x,y)=(a+_q c^bx,b+_py)##
There is also an explanation of what c is but very little else.
It doesn't even explain what operation adjacency represents, eq., ##c^bx##.
Then I am asked to prove that ##(a,b)^-1=(-c^{-b}a,b)##.
I wasn't able to solve it based on the skimpy material in the book.
I searched all over the internet and there is nothing about this formula.
Semidirect products are always defined in a totally different way.
Can anyone point to some examples of using this formula?
It probably won't do any good to explain the theory to me.
I work better the other way around.
When I understand how to do it, then I can understand the theory.
 

Answers and Replies

  • #2
fresh_42
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The natural guess is that it stands for conjugation: ##c^b(x)=b^{-1}xb## in general, or ##c^b(x)=-b+x+b## in this case. But since ##\mathbb{Z}_p,\mathbb{Z}_q## are both Abelian groups, every semidirect product is automatically direct, because the conjugation ##c## reduces to the identity map: ##c^b(x)=-b+x+b=x.##
 
  • #3
jstrunk
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Here is what the book says about c.
Let 1<c<q be minimal in ##Z^{*}_{q}## with o(c)=p. Observe that since o(c)=p we have ##c^p=1 \in Z^{*}_{q}##
and it follows that ##c^k=c^{kmodp} \in Z^{*}_{q}##.

The formula doesn't even make sense in the second term, which doesn't involve c.
I am asked to verify that ##(a,b)^{-1}=(-c^{-b}a,b)## applying the formula ##(a,b)*(x,y)=(a+_q c^bx,b+_py)##.
The identity is (0,0) so I need to show that ##(a,b)*(-c^{-b}a,b)=(0,0)## and ##(-c^{-b}a,b)*(a,b)=(0,0)##.
For the second term, that means ##b+_pb=0## for all primes p and ##b \in Z_p##.
That doesn't make sense. ##b+_p{-b}=0## would make more sense.
Maybe there is a typo in the book?
I was hoping someone would recognize these formulas and be able to point me to a better explanation or examples.
 
  • #4
fresh_42
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The crucial condition is that ##p|(q-1)##. This means ##q-1=a\cdot p## or ##q=ap+1##. We have for all elements ##x\in \{1,\ldots,q-1\}=\mathbb{Z}_q^*## that ##(x^a)^p=x^{ap}=x^{q-1}=1## because for every group ##G## we have ##x^{|G|}=1.## Thus we can choose ##c## as the smallest among those whose order is exactly ##p##, i.e. the minimum of all ##x^a##. And ##c^p=1 \in \mathbb{Z}^*_q.##

Next let ##k=-a\cdot p + r## with ##0\leq r < p##. Then ##c^r=c^{k+ap}=c^k\cdot c^{ap}=c^k\cdot (c^p)^a=c^k,## which means ##c^k=c^r=c^{k \mod p}## and of course it is still an element of ##\mathbb{Z}_q^*##.

The problem reads as is in this case: ##(a,b)*(x,y)=(a+_q c^bx,b+_py)##

##*## defines a binary operation. Whether this is a group operation isn't obvious here.
##+_p## and ##+_q## denote the addition in resp. ##\mathbb{Z}_p, \mathbb{Z}_q##
##c^b## is the multiplication in ##\mathbb{Z}^*_q##: ##\underbrace{c\cdot c\cdot \ldots\cdot c}_{b\text{ times}}##
Since ##Z_q^*\subseteq \mathbb{Z_q}## as a set, we can define ##c^b\cdot x## as multiplication in ##\mathbb{Z}_q^*## if ##q\nmid x## and ##c^bx=0## if ##q|x##.

So finally we have a well-defined binary operation ##*## on the ##\mathbb{Z}_q \times \mathbb{Z}_p##. Which properties this operation has isn't obvious. You will have to check.
 

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