Theorem: Groups of Order 21

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 $p q$ 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:
{{begin-eqn}}
{{eqn | l = x
      | r = y^3 x y^{-3}
      | c = 
}}
{{eqn | r = y^2 \paren {y x y^{-1} } y^{-2}
      | c = {{Group-axiom|1}}
}}
{{eqn | r = y^2 x^i y^2
      | c = as $y x y^{-1} = x^i$
}}
{{eqn | r = \paren {y^2 x y^{-2} }^i
      | c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = \paren {y \paren {y x y^{-1} } y^{-1} }^i
      | c = {{Group-axiom|1}}
}}
{{eqn | r = \paren {y x^i y^{-1} }^i
      | c = as $y x y^{-1} = x^i$
}}
{{eqn | r = y \paren {x^i}^i y^{-1}
      | c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = y \paren {x^{i^2} } y^{-1}
      | c = Powers of Group Elements
}}
{{eqn | r = \paren {y x y^{-1} }^{i^2}
      | c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = \paren {x^i}^{i^2}
      | c = as $y x y^{-1} = x^i$
}}
{{eqn | r = x^{i^3}
      | c = Powers of Group Elements
}}
{{end-eqn}}
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.
{{begin-eqn}}
{{eqn | ll= i = 0:
      | l = 0^3 - 1
      | r = -1
      | c = 
}}
{{eqn | o = \equiv
      | r = 6
      | rr= \pmod 7
      | c = so $0$ is not a possible value of $i$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ll= i = 1:
      | l = 1^3 - 1
      | r = 0
      | c = 
}}
{{eqn | o = \equiv
      | r = 0
      | rr= \pmod 7
      | c = so $1$ is a possible value of $i$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ll= i = 2:
      | l = 2^3 - 1
      | r = 7
      | c = 
}}
{{eqn | o = \equiv
      | r = 0
      | rr= \pmod 7
      | c = so $2$ is a possible value of $i$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ll= i = 3:
      | l = 3^3 - 1
      | r = 26
      | c = 
}}
{{eqn | o = \equiv
      | r = 5
      | rr= \pmod 7
      | c = so $3$ is not a possible value of $i$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ll= i = 4:
      | l = 4^3 - 1
      | r = 63
      | c = 
}}
{{eqn | o = \equiv
      | r = 0
      | rr= \pmod 7
      | c = so $4$ is a possible value of $i$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ll= i = 5:
      | l = 5^3 - 1
      | r = 124
      | c = 
}}
{{eqn | o = \equiv
      | r = 5
      | rr= \pmod 7
      | c = so $5$ is not a possible value of $i$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ll= i = 6:
      | l = 6^3 - 1
      | r = 215
      | c = 
}}
{{eqn | o = \equiv
      | r = 5
      | rr= \pmod 7
      | c = so $6$ is not a possible value of $i$
}}
{{end-eqn}}
Thus $i \bmod 7 \in \set {1, 2, 4}$.
{{qed|lemma}}
Suppose $i \equiv 1 \pmod 7$.
Then:
{{begin-eqn}}
{{eqn | l = y x y^{-1}
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y x
      | r = x y
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \paren {x y}^3
      | r = x^3 y^3
      | c = Power of Product of Commutative Elements in Group
}}
{{eqn | r = x^3
      | c = as $y^3 = e$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \paren {x y}^7
      | r = x^7 y^7
      | c = Power of Product of Commutative Elements in Group
}}
{{eqn | r = y^7
      | c = as $x^7 = e$
}}
{{eqn | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^{21}
      | r = y^3
      | c = 
}}
{{eqn | r = e
      | c = 
}}
{{end-eqn}}
It follows that:
:$\order {x y} = 21$
where $\order {x y}$ denotes the order of $x y$.
Thus $G$ is cyclic.
{{qed|lemma}}
Suppose that $i \equiv 2 \pmod 7$.
Thus, let $y$ be an element of order $3$ for which $y x y^{-1} = x^2$.
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:
:$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2 }$
{{qed}}