Theorem: Groups of Order 30/C 5 x D 3

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:
{{begin-eqn}}
{{eqn | l = y z y^{-1}
      | r = y x^3 y^{-1}
      | c = 
}}
{{eqn | r = \paren {y x y^{-1} }^3
      | c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = \paren {x^{11} }^3
      | c = from Group Presentation
}}
{{eqn | r = x^{33}
      | c = Powers of Group Elements: Product of Indices
}}
{{eqn | r = x^{15} x^{15} x^3
      | c = Powers of Group Elements: Sum of Indices
}}
{{eqn | r = e \cdot e \cdot x^3
      | c = from Group Presentation
}}
{{eqn | r = x^3
      | c = {{Group-axiom|2}}
}}
{{eqn | r = z
      | c = Definition of $z$
}}
{{eqn | ll= \leadsto
      | l = y z
      | r = z y
      | c = Product of both sides with $y$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = y x^5 y^{-1}
      | r = \paren {y x y^{-1} }^5
      | c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = \paren {x^{11} }^5
      | c = from Group Presentation
}}
{{eqn | r = x^{55}
      | c = Powers of Group Elements: Product of Indices
}}
{{eqn | r = x^{15} x^{15} x^{15} x^{10}
      | c = Powers of Group Elements: Sum of Indices
}}
{{eqn | r = e \cdot e \cdot e \cdot x^{10}
      | c = from Group Presentation
}}
{{eqn | r = x^{10}
      | c = {{Group-axiom|2}}
}}
{{eqn | r = x^{-5}
      | c = as $x^{15} = e$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = x w x^{-1}
      | r = x x^5 x^{-1}
      | c = Definition of $w$
}}
{{eqn | r = x^5
      | c = 
}}
{{eqn | r = w
      | c = 
}}
{{end-eqn}}
demonstrating that $x$ is conjugate to $w$.
Then:
{{begin-eqn}}
{{eqn | l = x y x^{-1}
      | r = x y^{-1} x^{-1}
      | c = as $y^2 = e$
}}
{{eqn | r = y x^{11} x^{-1}
      | c = from Group Presentation: $y x y^{-1} = x^{11}$
}}
{{eqn | r = y x^{10}
      | c = 
}}
{{eqn | r = y w^2
      | c = 
}}
{{eqn | o = \in
      | r = K
      | c = 
}}
{{end-eqn}}
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:
:$G = C_5 \times D_3$
{{qed}}