# What am I missing here? Trying to learn the semidirect product

I'm trying to understand the concept of the semidirect product of two groups.

An application is this answer to a question, where for $$p>q$$ primes, $$q \mid p-1$$, there is a non-abelian group of order $$pq$$ isomorphic to

$$C_p\rtimes_{\phi} C_q$$

for some homomorphism $$\phi: C_q\to\mathrm{Aut}(C_p)\cong C_{p-1}.$$

I'm trying to understand what this homomorphism looks like.

The answer states the following at point 2:

(Note the author defines $$P=\langle x\mid x^p=1\rangle$$ and $$Q=\langle y\mid y^q=1\rangle$$.)

Now, since $$Q=\langle y\rangle$$ normalizes $$P=\langle x\rangle$$, the map $$\phi_k:P\to P$$ given by $$\phi_k(x)=y^kxy^{-k}$$ is well defined. Moreover, it is clearly an automorphism with inverse $$\phi_{-k}$$. Finally, since $$\phi_{k}\phi_j=\phi_{k+j}$$, the map $$y^k\mapsto\phi_k$$ defines a homomorphism $$\phi:Q\to \mathrm{Aut}(P)$$.

Here's where I'm a bit confused - something I'm doing from here on must be wrong, but I'm not sure what. Let's represent $$C_p$$ and $$C_q$$ as addition modulo $$p$$ and $$q$$, thus the $$pq$$ elements of $$C_p\rtimes_{\phi} C_q$$ will be the tuples of the form $$(a, b)$$ for $$a =0, ..., p-1$$, $$b=0, ..., q-1$$.

In the equation $$\phi_k(x)=y^kxy^{-k}$$

$$y \in C_q$$ so it'll be of the form $$(0, m)$$ in the semidirect product, $$x \in C_p$$ will be of the form $$(n, 0)$$. Thus $$y^kxy^{-k}=(0, mk) \circ (n, 0) \circ(0,-mk)$$ where $$\circ$$ is the group operation of the semidirect product.

First, I thought this was just like addition on the tuples, and so I will always get $$(n, 0)=x$$. This interpretation must be wrong, since then $$\phi_k$$ would be the trivial homomorphism, and we just get the abelian direct product.

But $$\circ$$ is defined as $$(n_1, h_1) \circ (n_2, h_2) = (n_1 \phi_{h_1}(n_2), h_1 h_2)$$, which uses what will be $$\phi_k$$ in its own definition... so what's going wrong? I don't get how I can actually calculate binary operations on the group.

• You've introduced $Q$ and $P$ without saying what they are. I suppose by context the intention is $Q=C_q$ and $P=C_p$, but if so then you should hit the edit button to make that clear. Jun 4 at 13:14
• @LeeMosher My bad. Yes, that is what they are. Added that in the question. Jun 4 at 13:15

One thing I see wrong is that despite choosing additive notation for $$C_p$$ and $$C_q$$, you then use multiplication, for example in the expression $$h_1 h_2$$. But this is a minor error. I suggest dumping additive notation, which you are free to do. Since $$C_p$$ and $$C_q$$ are being used as subgroups of a semidirect product which is not going to be abelian, stick with the multiplicative notation $$C_p=\langle x \rangle$$ and $$C_q = \langle y \rangle$$. Also, I'm going to ignore the $$P,Q$$ notation and other doubled up notations. Why confuse yourself by doubling up on notations when one single notation will do?
Regarding $$\phi_k$$, the expression $$\phi_k(y) = y^k x y^{-k}$$ is internal to the semidirect product. As you say, this does not make sense in giving an external definition for a homomorphism $$C_q \mapsto \text{Aut}(C_p)$$.
So, let's just give an external definition of a homomorphism $$\Phi : C_q \to \text{Aut}(C_p)$$.
Start with the fact that the generator $$y \in C_q$$ has order $$q$$. Our job is to choose $$\Phi(y) \in \text{Aut}(C_p)$$ to be a nontrivial automorphism of the group $$C_p$$ such that the order of $$\Phi(y)$$ in the group $$\text{Aut}(C_p)$$ is equal to $$q$$ (making the choice so that the order divides $$q$$ would be sufficient; but we can actually make it equal to $$q$$). Once that job is done, the formula $$\Phi(y^k) = \underbrace{\Phi(y) \circ \ldots \circ \Phi(y)}_{\text{k times}}$$ becomes a well-defined homomorphism, using the composition operation $$\circ$$ which is the group operation on $$\text{Aut}(C_p)$$.
The fact that one may choose $$\Phi(y)$$ in this manner follows by combining the hypothesis $$q \mid p-1$$ with the theorem that $$\text{Aut}(C_p)$$ is a cyclic group of order $$p-1$$.
Now, you stated that you want to know what the homomorphism $$C_q \mapsto \text{Aut}(C_p)$$ looks like. So it's possible that this answer so far is not enough for you: perhaps you really want to know what an order $$q$$ element of the group $$\text{Aut}(C_p)$$ looks like. For that purpose, I would say that you should go read the proof of that theorem.