Groups of Order 30/C 5 x D 3

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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:

$G = C_5 \times D_3$

$\blacksquare$


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