Metacyclic Group – Exploring Semidirect and Metacyclic Groups in Group Theory

finite-groupsgroup-theorysemidirect-product

Let $G$ and $H$ be groups and $\theta : H \to\operatorname{Aut} G$ a homomorphism. Define $G\times_{\theta}H$ is called the semidirect product of $G$ and $H$.

Let $C_{p}=\langle a\rangle$ and $C_{q}=\langle b\rangle$ be (multiplicative) cyclic groups of prime orders $p$ and $q$ respectively such that $p > q$ and $q\mid p — 1$.

a. The map $\alpha:C_{p}\to C_{p}$ given by $a^{i}\mapsto a^{si}$ is an automorphism.

b. The map $\theta:C_{q}\to Aut C_{q }$ given by $\theta(b^{i}) =\alpha^{i}$ ($\alpha$ as in part (a)) is a homomorphism ($\alpha^{i} = I_{C_{p}})$.

c. If we write $a$ for $(a,e)$ and $b$ for $(e,b)$, then the group $C_{p}\times_{\theta} C_{g}$ is a group of order $pq$, generated by $a$ and $b$ subject to the relations:
$|a|=p$, $|b| = q$, $ba = a^{s}b$, where $s\not\equiv 1\pmod p$, and $s^{q}\equiv 1\pmod p$. The group $C_{p} \times_{\theta} C_{q}$ is called the metacyclic group.

I have tried to solve it, the a, since $C_{p}=\langle a \rangle=\lbrace a^{p}\mid\text{$p$ is prime}\rbrace$, hence for some $s\in \mathbb{Z}$, $(s,p)=1$, in this case $\alpha^{s}$ is also a generator of $C_{p}$. Now for some $m\in \mathbb{Z}$ imples $s^{m}\equiv1\pmod p$, the map $\alpha:C_{p}\to C_{p}$ defined an automorphism.
Calculated $\alpha^{m}(\alpha^{i})=\alpha^{m-1}(\alpha^{si}) \cdots =\alpha^{s^{m}i}=\alpha^{i}=e$.

For b, I tried use theorem Dyck, but I'm not sure.

I would like to know how to solve it or any suggestions, I appreciate.

Best Answer

Let $q | p-1$ and $C_p = \langle a \rangle$ and $C_q = \langle b \rangle$.

For the semidirect product $C_p \rtimes_\theta C_q$ we will need to define a group homomorphism $\theta : C_q \to \operatorname{Aut}(C_p)$.

We will have a group of order $pq$ and $C_p \lhd C_p \rtimes_\theta C_q$

First $\operatorname{Aut}(C_p)$

$\alpha : C_p \to C_p$

$\alpha(a^i) = a^{si}$

will be an automorphism, that is a group isomorphism from $C_p$ to $C_p$, that is a group homomorphism that is also a bijection.

We can show it is a group homomorphism:

  • $\alpha(a^i a^j) = a^{s(i+j)}$
  • $\alpha(a^i)\alpha(a^j) = a^{si}a^{sj}$

and these are equal so it is.

And it will be a bijection if multiplication by $s$ is invertible mod $p$.

$\theta : C_q \to \operatorname{Aut}(C_p)$

$\theta(b^i) = \alpha^i$

We will show this is a group homomorphism:

  • $\theta(b^i b^j) = \alpha^{i+j}$ applying to $a^k$: $a^{s^{i+j} k}$.
  • $\theta(b^i) \circ \theta(b^j) = \alpha^{i} \circ \alpha^{i}$ applying to $a^k$: $\alpha^{i}(a^{s^j k}) = a^{s^i s^j k}$

and these are equal so this is a valid group homomorphism.

Detail on $s$:

From $\alpha$ being invertible we require that $s$ is a unit mod $p$.

From $\theta$ being a group homomorphism from $C_q$ (i.e. $\theta(b^q) = \theta(1)$) we require that $\alpha^q = 1$. So we need $s^q \equiv 1 \pmod p$.

Now we will always have $s^{p-1} \equiv 1 \pmod p$ so we can take a primitive root $r$ and notice $(s^{\frac{p-1}{q}})^q \equiv 1 \pmod p$ so we find an $s$ by raising $r$ to the power $(p-1)/q$.

In general the semidirect product has multiplication operation as follows ($b$ is a general element, not the generator for $C_q$ for the next line only):

$$(b,g)(c,h) = (b \theta(g)(c), gh)$$

So in our case

$$ba = (1,b)(a,1) = (\theta(b)(a),b) = (\alpha(a),b) = (a^{s}, b) = a^{s} b$$

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