# Groups of Order 30/Lemma

## 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:

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

Then:

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

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

$\blacksquare$