# 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:

\(\displaystyle x\) | \(=\) | \(\displaystyle y^3 x y^{-3}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y^2 \paren {y x y^{-1} } y^{-2}\) | Group Axiom $\text G 1$: Associativity | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y^2 x^i y^2\) | as $y x y^{-1} = x^i$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {y^2 x y^{-2} }^i\) | Power of Conjugate equals Conjugate of Power | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {y \paren {y x y^{-1} } y^{-1} }^i\) | Group Axiom $\text G 1$: Associativity | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {y x^i y^{-1} }^i\) | as $y x y^{-1} = x^i$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y \paren {x^i}^i y^{-1}\) | Power of Conjugate equals Conjugate of Power | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y \paren {x^{i^2} } y^{-1}\) | Powers of Group Elements | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {y x y^{-1} }^{i^2}\) | Power of Conjugate equals Conjugate of Power | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {x^i}^{i^2}\) | as $y x y^{-1} = x^i$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{i^3}\) | 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.

\(\displaystyle i = 0: \ \ \) | \(\displaystyle 0^3 - 1\) | \(=\) | \(\displaystyle -1\) | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle 6\) | \(\displaystyle \pmod 7\) | so $0$ is not a possible value of $i$ |

\(\displaystyle i = 1: \ \ \) | \(\displaystyle 1^3 - 1\) | \(=\) | \(\displaystyle 0\) | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle 0\) | \(\displaystyle \pmod 7\) | so $1$ is a possible value of $i$ |

\(\displaystyle i = 2: \ \ \) | \(\displaystyle 2^3 - 1\) | \(=\) | \(\displaystyle 7\) | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle 0\) | \(\displaystyle \pmod 7\) | so $2$ is a possible value of $i$ |

\(\displaystyle i = 3: \ \ \) | \(\displaystyle 3^3 - 1\) | \(=\) | \(\displaystyle 26\) | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle 5\) | \(\displaystyle \pmod 7\) | so $3$ is not a possible value of $i$ |

\(\displaystyle i = 4: \ \ \) | \(\displaystyle 4^3 - 1\) | \(=\) | \(\displaystyle 63\) | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle 0\) | \(\displaystyle \pmod 7\) | so $4$ is a possible value of $i$ |

\(\displaystyle i = 5: \ \ \) | \(\displaystyle 5^3 - 1\) | \(=\) | \(\displaystyle 124\) | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle 5\) | \(\displaystyle \pmod 7\) | so $5$ is not a possible value of $i$ |

\(\displaystyle i = 6: \ \ \) | \(\displaystyle 6^3 - 1\) | \(=\) | \(\displaystyle 215\) | ||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle 5\) | \(\displaystyle \pmod 7\) | 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:

\(\displaystyle y x y^{-1}\) | \(=\) | \(\displaystyle x\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle y x\) | \(=\) | \(\displaystyle x y\) |

Hence:

\(\displaystyle \paren {x y}^3\) | \(=\) | \(\displaystyle x^3 y^3\) | Power of Product of Commutative Elements in Group | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^3\) | as $y^3 = e$ |

and:

\(\displaystyle \paren {x y}^7\) | \(=\) | \(\displaystyle x^7 y^7\) | Power of Product of Commutative Elements in Group | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y^7\) | as $x^7 = e$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x^{21}\) | \(=\) | \(\displaystyle y^3\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e\) |

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

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: $(2)$ Groups of order $21$