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 $y x y^{-1} = x^2$.
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:
- $\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2 }$
$\blacksquare$
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
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: $(2)$ Groups of order $21$