Groups of Order 21

Theorem
There exist exactly $2$ groups of order $21$, up to isomorphism:


 * $(1): \quad C_{21}$, the cyclic group of order $21$


 * $(2): \quad$ the group whose group presentation is:
 * $\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2}$

Proof
Let $G$ be of order $21$.

From Group of Order pq has Normal Sylow p-Subgroup, $G$ has exactly one Sylow $7$-subgroup, which is normal.

Let this Sylow $7$-subgroup of $G$ be denoted $P = \gen {x: x^7 = 1}$.

From the First Sylow Theorem, $G$ also has at least one Sylow $3$-subgroup.

Thus there exists $y \in G$ of order $3$.

As $P$ is normal:
 * $y x y^{-1} = x^i$

for some $i \in \set {0, 1, \ldots, 6}$.

Thus:

So $x^1 = x^{i^3}$ and so:
 * $i^3 \equiv 1 \pmod 7$

and so:
 * $7 \divides \paren {i^3 - 1}$

where $\divides$ indicates divisibility.

Let us consider the $7$ possible values of $i$ in turn.

Thus $i \bmod 7 \in \set {1, 2, 4}$.

Suppose $i \equiv 1 \pmod 7$.

Then:

Hence:

and:

It follows that:
 * $\order {x y} = 21$

where $\order {x y}$ denotes the order of $x y$.

Thus $G$ is cyclic.

Suppose that $i \equiv 2 \pmod 7$.

Thus, let $y$ be an element of order $3$ for which $u x y^{-1} = x^2$.

Then $z = y^2$ is an element of order $3$ for which $z x z^{-1} = x^4$.

Thus the group as defined here where $i = 2$ is isomorphic to the group as defined here where $i = 4$.

Thus, apart from $C_{21}$, the other group of order $21$ has the group presentation:


 * $\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2 }$