# 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:

$G = C_3 \times D_5$

$\blacksquare$