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
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: $(5)$ Groups of order $30$: Proposition $12.6$