Groups of Order 21

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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 $\Z_7$:

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


Sources