# Groups of Order 30/C 5 x D 3

## Theorem

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

Let $G$ have the group presentation:

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

Then $G$ is isomorphic to the group direct product of the cyclic group $C_5$ and the dihedral group $D_3$:

- $G \cong C_5 \times D_3$

## Proof

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:

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

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

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

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

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

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

\(\displaystyle \) | \(=\) | \(\displaystyle x^3\) | Group Axiom $G \, 2$: Identity | ||||||||||

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

\(\displaystyle \leadsto \ \ \) | \(\displaystyle y z\) | \(=\) | \(\displaystyle z y\) | 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:

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

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

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

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

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

\(\displaystyle \) | \(=\) | \(\displaystyle x^{10}\) | Group Axiom $G \, 2$: Identity | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{-5}\) | 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:

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

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

\(\displaystyle \) | \(=\) | \(\displaystyle w\) |

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

Then:

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

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

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

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

\(\displaystyle \) | \(\in\) | \(\displaystyle K\) |

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:

- $G = C_5 \times D_3$

$\blacksquare$

## Sources

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