We have two possibilities, I know that one of the possibilities is the cyclic group$\frac{\Bbb{Z}}{21\Bbb{Z}}$. The other possibility as shown below with Sylow's theorems is $\Bbb{Z}_7 \times \Bbb{Z}_3$, where I have trouble is in the second case. Where there can be $7$ Sylow-3 Subgroups. Why is the second case isomorphic to $\Bbb{Z}_7 \times \Bbb{Z}_3$
We have two sylow subgroups of orders 7 and 3. Let $n_3$ and $n_7$ denote the number of sylow subgroups for 3 and 7, respectively.
$n_7 \equiv 1 \mod 7$ and $n_7 | 3 \implies n_7 = 1$
$n_3 \equiv 1 \mod 3$ and $n_3 | 7 \implies n_3 = 1, 7$
Let $P_3 \cong \Bbb{Z}_3 $ and $P_7 \cong \Bbb{Z}_7$. Since $P_7$ is always normal in G, we know that $G \cong \Bbb{Z}_7 \times \Bbb{Z}_3$.
Case 1:
Let $n_3 = 1$. Then we know that $P_3$ is also normal, and so $G \cong \Bbb{Z}_7 \times \Bbb{Z}_3$.
Case 2:
Let $n_3=7$. We have $: \Bbb{Z}_3 \rightarrow \Bbb{Z}_7 \cong \Bbb{Z}_6$
Now how do I construct this group called the Frobenious group? I do not know semi products, any help would be appreciated.
Best Answer
Here is a way to construct the nonabelian group of order $21$ without referring to semidirect products.
Let $F$ be the field with $7$ elements, which we will identify with the set $\{0,1,2,3,4,5,6\}$. Note that the elements $H = \{1,2,4\}$ form a subgroup of the units of $F$ of order $3$.
Now define $G = \{f: F\to F\mid f(x) = ax + b,a\in H, b\in F\}$.
This is a group under composition of functions, and it is easy to check that it has order $21$ and is nonabelian.
In general, given a finite field of order $q$, we can make a construction as above using any non-trivial subgroup of the multiplicative group for $H$, producing nonabelian groups of order $qm$ for any non-trivial divisor $m$ of $q-1$ (m = |H|).