Groups of Order 30/Lemma

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Theorem

Let $G$ be a group of order $30$.

Then $G$ is one of the following:

The cyclic group $C_{30}$
The dihedral group $D_{15}$
Isomorphic to one of:
$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$


Proof

By Group of Order 30 has Normal Cyclic Subgroup of Order 15, $G$ has a normal subgroup of order $15$ which is cyclic.

Let this normal cyclic order $15$ subgroup be denoted $N$:

$N = \gen x$


Let $y$ be the generator for any Sylow $2$-subgroup of $G$.

Then:

\(\ds y x y^{-1}\) \(\in\) \(\ds N\) as $N$ is normal
\(\ds \leadsto \ \ \) \(\ds y x y^{-1}\) \(=\) \(\ds x^i\) for some $i \in \Z_{\ge 0}$

Then:

\(\ds x\) \(=\) \(\ds y^2 x y^{-2}\) as $y^2 = e$
\(\ds \) \(=\) \(\ds y \paren {y x y^{-1} } y^{-1}\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds y x^i y^{-1}\) as $y x y^{-1} = x^i$
\(\ds \) \(=\) \(\ds \paren {y x y^{-1} }^i\) Power of Conjugate equals Conjugate of Power
\(\ds \) \(=\) \(\ds \paren {x^i}^i\) as $y x y^{-1} = x^i$
\(\ds \) \(=\) \(\ds x^{i^2}\) Powers of Group Elements

and so:

$i^2 - 1 \equiv 0 \pmod {15}$

Investigating the powers of $i$, case by case, searching for those which satisfy this congruence, yields:

$i \in \set {1, 4, 11, 14}$


The case $i \equiv 1 \pmod {15}$ leads to the cyclic group $C_{30}$.


The case where $i \equiv {14} \equiv {-1} \pmod {15}$ leads to the dihedral group $D_{15}$.


The other two cases lead to:

$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$

$\blacksquare$


Sources

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