# 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:

$G = C_5 \times D_3$

$\blacksquare$