# Groups of Order 30

## Theorem

Let $G$ be a group of order $30$.

Then $G$ is one of the following:

- The cyclic group $C_{30}$

- The dihedral group $D_{15}$

- The group direct product $C_5 \times D_3$

- The group direct product $C_3 \times D_5$

## Proof

First we introduce a lemma:

### Lemma

Let $G$ be a group of order $30$.

Then $G$ is one of the following:

- The cyclic group $C_{30}$

- The dihedral group $D_{15}$

- Isomorphic to one of:

- $\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$

- $\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$

$\Box$

From the lemma, it remains to be shown that the group presentations:

- $\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$

- $\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$

give the groups $C_5 \times D_3$ and $C_3 \times D_5$.

### Direct Product $C_5 \times D_3$

Let $G$ be defined by its group presentation:

- $G = \gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$

Let $z$ denote $x^3$.

Then:

\(\ds y z y^{-1}\) | \(=\) | \(\ds y x^3 y^{-1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {y x y^{-1} }^3\) | Power of Conjugate equals Conjugate of Power | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {x^{11} }^3\) | from Group Presentation | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{33}\) | Powers of Group Elements: Product of Indices | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{15} x^{15} x^3\) | Powers of Group Elements: Sum of Indices | |||||||||||

\(\ds \) | \(=\) | \(\ds e \cdot e \cdot x^3\) | from Group Presentation | |||||||||||

\(\ds \) | \(=\) | \(\ds x^3\) | Group Axiom $\text G 2$: Existence of Identity Element | |||||||||||

\(\ds \) | \(=\) | \(\ds z\) | Definition of $z$ | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds y z\) | \(=\) | \(\ds z y\) | Product of both sides with $y$ |

So $z$ commutes with $y$.

As $z$ is a power of $x$, $z$ also commutes with $x$.

Hence by definition of center:

- $z \in \map Z G$

It follows that $\gen z$ is a normal subgroup of order $5$.

Let $K$ be the subgroup of $G$ generated by $x^5$ and $y$.

Note that:

\(\ds y x^5 y^{-1}\) | \(=\) | \(\ds \paren {y x y^{-1} }^5\) | Power of Conjugate equals Conjugate of Power | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {x^{11} }^5\) | from Group Presentation | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{55}\) | Powers of Group Elements: Product of Indices | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{15} x^{15} x^{15} x^{10}\) | Powers of Group Elements: Sum of Indices | |||||||||||

\(\ds \) | \(=\) | \(\ds e \cdot e \cdot e \cdot x^{10}\) | from Group Presentation | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{10}\) | Group Axiom $\text G 2$: Existence of Identity Element | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{-5}\) | as $x^{15} = e$ |

Hence the generator of $K$ satisfies:

- $\paren{x^5}^3 = e = y^2$

and:

- $y x^5 y^{-1} = x^{-5}$

Let $w := x^5$.

Then $K$ is generated by $w$ and $y$ where:

- $w^3 = 1 = y^2$

and:

- $w y = y w^2 = y w^{-1}$

and it is seen that $K$ is isomorphic to the dihedral group $D_3$.

It is now to be shown that $G$ is an internal group direct product of $K$ and $\gen z$.

We have that $K \cap \gen z = \set e$.

From the Internal Direct Product Theorem, we need to prove only that $K$ is a normal subgroup of $G$.

We have that:

- $\order K = 6$

where $\order K$ denotes the order of $K$.

We also have that $K$ is a subgroup of its normalizer $\map {N_G} K$.

Hence by Lagrange's Theorem:

- $6 \divides \order {\map {N_G} K}$

where $\divides$ denotes divisibility.

Again by Lagrange's Theorem:

- $\order {\map {N_G} K} \divides 30$

We have:

\(\ds x w x^{-1}\) | \(=\) | \(\ds x x^5 x^{-1}\) | Definition of $w$ | |||||||||||

\(\ds \) | \(=\) | \(\ds x^5\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds w\) |

demonstrating that $x$ is conjugate to $w$.

Then:

\(\ds x y x^{-1}\) | \(=\) | \(\ds x y^{-1} x^{-1}\) | as $y^2 = e$ | |||||||||||

\(\ds \) | \(=\) | \(\ds y x^{11} x^{-1}\) | from Group Presentation: $y x y^{-1} = x^{11}$ | |||||||||||

\(\ds \) | \(=\) | \(\ds y x^{10}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds y w^2\) | ||||||||||||

\(\ds \) | \(\in\) | \(\ds K\) |

demonstrating that $x$ is conjugate to $y$.

Thus $x \in \map {N_G} K$ and so $\order {\map {N_G} K} > 6$.

As $6 \divides \order {\map {N_G} K}$ and $\order {\map {N_G} K} \divides 30$, it follows that:

- $\map {N_G} K = G$

and so $K$ is normal in $G$.

Thus:

- $K$ and $\gen z$ are normal in $G$
- $K \cap \gen z = \set e$
- $K \gen z = G$

and it therefore follows from the Internal Direct Product Theorem that:

- $G = C_5 \times D_3$

$\blacksquare$

### Direct Product $C_3 \times D_5$

Let $G$ be defined by its group presentation:

- $G = \gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$

Let $z$ denote $x^5$.

Then:

\(\ds y z y^{-1}\) | \(=\) | \(\ds y x^5 y^{-1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {y x y^{-1} }^5\) | Power of Conjugate equals Conjugate of Power | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {x^4}^5\) | from Group Presentation | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{20}\) | Powers of Group Elements: Product of Indices | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{15} x^5\) | Powers of Group Elements: Sum of Indices | |||||||||||

\(\ds \) | \(=\) | \(\ds e \cdot x^5\) | from Group Presentation | |||||||||||

\(\ds \) | \(=\) | \(\ds x^5\) | Group Axiom $\text G 2$: Existence of Identity Element | |||||||||||

\(\ds \) | \(=\) | \(\ds z\) | Definition of $z$ | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds y z\) | \(=\) | \(\ds z y\) | Product of both sides with $y$ |

So $z$ commutes with $y$.

As $z$ is a power of $x$, $z$ also commutes with $x$.

Hence by definition of center:

- $z \in \map Z G$

It follows that $\gen z$ is a normal subgroup of order $3$.

Let $N$ be the subgroup of $G$ generated by $x^3$ and $y$.

Note that:

\(\ds y x^3 y^{-1}\) | \(=\) | \(\ds \paren {y x y^{-1} }^3\) | Power of Conjugate equals Conjugate of Power | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {x^4}^5\) | from Group Presentation | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{12}\) | Powers of Group Elements: Product of Indices | |||||||||||

\(\ds \) | \(=\) | \(\ds x^{-3}\) | as $x^{15} = e$ |

Hence the generator of $N$ satisfies:

- $\paren{x^3}^5 = e = y^2$

and:

- $y x^3 y^{-1} = x^{-3}$

Let $w := x^3$.

Then $N$ is generated by $w$ and $y$ where:

- $w^5 = 1 = y^2$

and:

- $w y = y w^4 = y w^{-1}$

and it is seen that $N$ is isomorphic to the dihedral group $D_5$.

It is now to be shown that $G$ is an internal group direct product of $N$ and $\gen z$.

We have that $N \cap \gen z = \set e$.

From the Internal Direct Product Theorem, we need to prove only that $N$ is a normal subgroup of $G$.

We have that:

- $\order N = 10$

where $\order N$ denotes the order of $N$.

We also have that $N$ is a subgroup of its normalizer $\map {N_G} N$.

Hence by Lagrange's Theorem:

- $10 \divides \order {\map {N_G} N}$

where $\divides$ denotes divisibility.

Again by Lagrange's Theorem:

- $\order {\map {N_G} N} \divides 30$

We have:

\(\ds x w x^{-1}\) | \(=\) | \(\ds x x^3 x^{-1}\) | Definition of $w$ | |||||||||||

\(\ds \) | \(=\) | \(\ds x^3\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds w\) |

demonstrating that $x$ is conjugate to $w$.

Then:

\(\ds x y x^{-1}\) | \(=\) | \(\ds x y^{-1} x^{-1}\) | as $y^2 = e$ | |||||||||||

\(\ds \) | \(=\) | \(\ds y x^4 x^{-1}\) | from Group Presentation: $y x y^{-1} = x^4$ | |||||||||||

\(\ds \) | \(=\) | \(\ds y x^3\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds y w\) | ||||||||||||

\(\ds \) | \(\in\) | \(\ds N\) |

demonstrating that $x$ is conjugate to $y$.

Thus $x \in \map {N_G} N$ and so $\order {\map {N_G} N} > 10$.

As $10 \divides \order {\map {N_G} N}$ and $\order {\map {N_G} N} \divides 30$, it follows that:

- $\map {N_G} N = G$

and so $N$ is normal in $G$.

Thus:

- $N$ and $\gen z$ are normal in $G$
- $N \cap \gen z = \set e$
- $N \gen z = G$

and it therefore follows from the Internal Direct Product Theorem that:

- $G = C_3 \times D_5$

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $13$: Direct products: Theorem $13.8$