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 = {{GroupAxiom|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 = {{GroupAxiom|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 $u x y^{-1} = x^2$. 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: :$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2 }$ {{qed}} === Matrix Representation of Non-Abelian Instance === {{:Groups of Order 21/Matrix Representation of Non-Abelian Instance}}