Examining the case when $p|q-1$ when $G$ is a group of $pq$ where $p$ and $q$ are primes such that $p

Question

I was studying Abstract Algebra by Dummit and Foote. In section 5.5 titled as "Semidirect Products" there was an example with the heading, Groups of order pq. The first paragraph of which dealt with showing that $G\cong \Bbb Z_{p}\times \Bbb Z_q$ if $p\nmid q-1.$

The next paragraph examines the case when $p|q-1.$

It goes like this:

Consider now the case when $p \mid q - 1$ and let $P = \langle y \rangle$. Since $\text{Aut}(Q)$ is cyclic it contains a unique subgroup of order $p$, say $\langle \gamma \rangle$, and any homomorphism $\varphi : P \to \text{Aut}(Q)$ must map $y$ to a power of $\gamma$. There are therefore $p$ homomorphisms $\varphi_i : P \to \text{Aut}(Q)$ given by $\varphi_i(y) = \gamma^i$, $0 \le i \le p - 1$. Since $\varphi_0$ is the trivial homomorphism, $Q \rtimes_{\varphi_0} P \cong Q \times P$ as before. Each $\varphi_i$ for $i \ne 0$ gives rise to a non-abelian group, $G_i$, of order $pq$. It is straightforward to check that these groups are all isomorphic because for each $\varphi_i$, $i > 0$, there is some generator $y_i$ of $P$ such that $\varphi_i(y_i) = \gamma$. Thus, up to a choice for the (arbitrary) generator of $P$, these semidirect products are all the same.

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I don't get how do they say that, "these groups are all isomorphic because for each $\varphi_i$, $i > 0$" ? Their reasoning seems somewhat too verbal. Moreover, I don't get their reasoning at all.

Answer 1

$\varphi_1(y^i) = \gamma^i$, so the mapping $x \mapsto x$, $y\mapsto y^i$ induces an isomorphism $Q \rtimes_{\varphi_i}P \to Q \rtimes_{\varphi_1}P$.

More precisely, the isomorphism $f: Q \rtimes_{\varphi_i}P \to Q \rtimes_{\varphi_1}P$ is defined by $f((x^a,y^b)):=(x^a,y^{bi})$ for $0 \le a < q$, $0 \le b < p$, where $Q = \langle x \rangle$, $P = \langle y \rangle$, and the exponent $bi$ in $y^{bi}$ needs to be reduced modulo $p$..

Answer 2

A nontrivial homomorphism $\varphi \colon \mathbb Z/p\mathbb Z\to \operatorname{Aut}(\mathbb Z/q\mathbb Z)$ is completely determined by $\varphi_i$, where $i$ is any generator of $\mathbb Z/p\mathbb Z$, namely any among $i=1,\dots,p-1$. In fact, by definition of generator, for every $k\in\mathbb Z/p\mathbb Z$, there's an integer $j$, $1\le j\le p-1$, such that $k=ji$; hence, $\varphi_k=$ $\varphi_{ji}=\varphi_{\underbrace{i+\dots+i}_{j\text{ times}}}=$ $\varphi_i^j$.

So, we are left to know what $\varphi_i$ is. Since the order of the image a group element under a homomorphism divides the order of the element, $\varphi_i$ must have order $p$, because $0$ solely lies in the trivial kernel on $\varphi$. Again as above, besides the necessary $\varphi_i(0)=0$, $\varphi_i$ is completely determined by $\varphi_i(1)$, and $\varphi_i(1):=k_i$ does the job, where $k_i$ is any of the $p-1$ elements of order $p$ in $(\mathbb Z/q\mathbb Z)^\times$.

Now, let: $$\varphi: i\mapsto (x\mapsto k_ix)$$ and $$\psi: i\mapsto (x\mapsto l_ix)$$ where $k_i,l_i$ are two $p$-th roots of unity of $(\mathbb Z/q\mathbb Z)^\times$. Since $k_i$ is a generator of the unique subgroup of $(\mathbb Z/q\mathbb Z)^\times$ of order $p$, there is a unique integer $m_i$ such that $k_i^{m_i}=l_i$. But then, for every $x\in\mathbb Z/q\mathbb Z$: \begin{alignat}{1} \psi_i(x) &= l_ix \\ &= k_i^{m_i}x \\ &= \varphi_i^{m_i}(x) \end{alignat} whence $\psi_i=\varphi_i^{m_i}$, namely $\psi_1^i=\varphi_1^{im_i\pmod p}$. Therefore, $\operatorname{im}\psi=\operatorname{im}\varphi$, and hence the two semidirect products are isomorphic.