# Weird Semidirect Product Formula

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• jstrunk
In summary, the author provides a semidirect product formula for groups ##Z_p## and ## Z_q## for primes p<q and p divides (q-1). There is very little information about what c is or what operation adjacency represents. The author is then asked to verify a property of the semidirect product, but is not able to do so based on the sparse information provided in the book.
jstrunk
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.
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.

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.##

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.

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.

## 1. What is the "Weird Semidirect Product Formula"?

The "Weird Semidirect Product Formula" is a mathematical formula used to compute the semidirect product of two groups, which is a way of combining two groups to create a new group.

## 2. How is the "Weird Semidirect Product Formula" different from other group product formulas?

The "Weird Semidirect Product Formula" is different from other group product formulas because it takes into account the action of one group on the other, rather than just combining the elements of the two groups.

## 3. What is the purpose of using the "Weird Semidirect Product Formula"?

The "Weird Semidirect Product Formula" is used to construct new groups with specific properties by combining two existing groups in a specific way.

## 4. Can the "Weird Semidirect Product Formula" be applied to any two groups?

No, the "Weird Semidirect Product Formula" can only be applied to groups that satisfy certain conditions, such as one group acting on the other by automorphisms.

## 5. How is the "Weird Semidirect Product Formula" used in real-world applications?

The "Weird Semidirect Product Formula" has applications in various fields such as physics, computer science, and cryptography. It is used to study symmetries and to construct new mathematical structures with specific properties.

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