Construct Group of Order 21 Without Semi Product

Question

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.

Answer 1

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|).

Answer 2

I like the construction Tobias has posted. Here is a more hands-on method.

The Sylow theorems tell us that a group of order $21$ has a normal subgroup of order $7$. Let this be generated by $a$, and let $b$ be an element of order $3$.

Then, because $a$ generates a normal subgroup, we have $bab^{-1}=a^r$ for some $1\le r \le 6$. If $r=1$ the group is abelian, so we assume $r\neq 1$.

Now note that $a=b^3ab^{-3}=b^2bab^{-1} \dots bab^{-1}b^{-2}$ where there are $r$ factors of form $bab^{-1}$ and this is equal to $b^2a^rb^{-2}=ba^{r^2}b^{-1}=a^{r^3}$ using a similar process.

This means we have $b^{r^3}=b$ so that $r^3\equiv 1 \bmod 7$ with $r\neq 1$

$2^3=8\equiv 1, 3^3=27\equiv -1, 4^3=64 \equiv 1, 5^3=125\equiv -1, 6^3=216 \equiv -1$

Hence we have two constructions $bab^{-1}=a^2$ and $bab^{-1}=a^4$. In the second of these, if we set $c=b^2$ we have $cac^{-1}=a^2$, so the groups are isomorphic, and there is only one new possibility.

So we have the possibility $a^7=b^3=1; bab^{-1}=a^2$. This last relation can be used to put every element in the form $a^pb^q$ - and there are $21$ distinct possibilities as required.

Now this construction is long-winded and seemingly relies on some coincidences - the semi-direct product construction organises the calculations, shows that they work, and is also rather more general than this simple case.