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:
- $G = C_3 \times D_5$
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $13$: Direct products: Theorem $13.8$