# 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 $p q$ 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:

 $\ds x$ $=$ $\ds y^3 x y^{-3}$ $\ds$ $=$ $\ds y^2 \paren {y x y^{-1} } y^{-2}$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds y^2 x^i y^2$ as $y x y^{-1} = x^i$ $\ds$ $=$ $\ds \paren {y^2 x y^{-2} }^i$ Power of Conjugate equals Conjugate of Power $\ds$ $=$ $\ds \paren {y \paren {y x y^{-1} } y^{-1} }^i$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds \paren {y x^i y^{-1} }^i$ as $y x y^{-1} = x^i$ $\ds$ $=$ $\ds y \paren {x^i}^i y^{-1}$ Power of Conjugate equals Conjugate of Power $\ds$ $=$ $\ds y \paren {x^{i^2} } y^{-1}$ Powers of Group Elements $\ds$ $=$ $\ds \paren {y x y^{-1} }^{i^2}$ Power of Conjugate equals Conjugate of Power $\ds$ $=$ $\ds \paren {x^i}^{i^2}$ as $y x y^{-1} = x^i$ $\ds$ $=$ $\ds x^{i^3}$ Powers of Group Elements

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.

 $\ds i = 0: \ \$ $\ds 0^3 - 1$ $=$ $\ds -1$ $\ds$ $\equiv$ $\ds 6$ $\ds \pmod 7$ so $0$ is not a possible value of $i$

 $\ds i = 1: \ \$ $\ds 1^3 - 1$ $=$ $\ds 0$ $\ds$ $\equiv$ $\ds 0$ $\ds \pmod 7$ so $1$ is a possible value of $i$

 $\ds i = 2: \ \$ $\ds 2^3 - 1$ $=$ $\ds 7$ $\ds$ $\equiv$ $\ds 0$ $\ds \pmod 7$ so $2$ is a possible value of $i$

 $\ds i = 3: \ \$ $\ds 3^3 - 1$ $=$ $\ds 26$ $\ds$ $\equiv$ $\ds 5$ $\ds \pmod 7$ so $3$ is not a possible value of $i$

 $\ds i = 4: \ \$ $\ds 4^3 - 1$ $=$ $\ds 63$ $\ds$ $\equiv$ $\ds 0$ $\ds \pmod 7$ so $4$ is a possible value of $i$

 $\ds i = 5: \ \$ $\ds 5^3 - 1$ $=$ $\ds 124$ $\ds$ $\equiv$ $\ds 5$ $\ds \pmod 7$ so $5$ is not a possible value of $i$

 $\ds i = 6: \ \$ $\ds 6^3 - 1$ $=$ $\ds 215$ $\ds$ $\equiv$ $\ds 5$ $\ds \pmod 7$ so $6$ is not a possible value of $i$

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

$\Box$

Suppose $i \equiv 1 \pmod 7$.

Then:

 $\ds y x y^{-1}$ $=$ $\ds x$ $\ds \leadsto \ \$ $\ds y x$ $=$ $\ds x y$

Hence:

 $\ds \paren {x y}^3$ $=$ $\ds x^3 y^3$ Power of Product of Commutative Elements in Group $\ds$ $=$ $\ds x^3$ as $y^3 = e$

and:

 $\ds \paren {x y}^7$ $=$ $\ds x^7 y^7$ Power of Product of Commutative Elements in Group $\ds$ $=$ $\ds y^7$ as $x^7 = e$ $\ds$ $=$ $\ds y$ $\ds \leadsto \ \$ $\ds x^{21}$ $=$ $\ds y^3$ $\ds$ $=$ $\ds e$

It follows that:

$\order {x y} = 21$

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

Thus $G$ is cyclic.

$\Box$

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:

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

$\blacksquare$

### Matrix Representation of Non-Abelian Instance

Let $G$ be the group of order $21$ whose group presentation is:

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

Then $G$ can be instantiated by the following pair of matrices over $\Z_7$:

$X = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \qquad Y = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$