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 $D3$:


 * $G \cong C_5 \times D_3$

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:

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:

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:

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

Then:

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:
 * $G = C_5 \times D_3$