Weird Semidirect Product Formula

[jstrunk]

TL;DR

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.

 

[fresh_42]

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: $c^b(x)=-b+x+b=x.$

 

[jstrunk]

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.

 

[fresh_42]

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.