A6 - subgroup tests

these are the full set of steps:

• identify the property $P$ that distinguishes the elements of $H$
• prove that identity has property $P$, ie. $H\ne \mathrm{\Phi }$
• assume that $a$ and $b$ have the property $P$
• use the assumption to show that $a{b}^{-1}$ has the property $P$

eg: let $G$ be an Abelian group with identity $e$, and $H=\{x\in G\mid x^2=e\}\le G$

firstly, $e^2=e⟹H\ne \mathrm{\Phi }$

let $a,b\in H⟹a^2=b^2=e$

$G$ is Abelian $⟹(a{b}^{-1}{)}^2=a{b}^{-1}a{b}^{-1}=a^2(b^{-1}{)}^2=a^2(b^2{)}^{-1}=e{e}^{-1}=e$

therefore, $a{b}^{-1}\in H$, and $H\le G$

• show that $ϵ\notin H$
• find $a\in H$ such that $a^{-1}\notin H$
• find $a,b\in H$ such that $ab\notin H$

$a\in ⟨a⟩⟹⟨a⟩\ne \mathrm{\Phi }$

$a^n,a^m\in ⟨a⟩$ then, $a^n(a^m{)}^{-1}=a^{n-m}\in ⟨a⟩$

therefore, $⟨a⟩\le G$

$⟨a⟩$ is called the cyclic subgroup of $G$ generated by $a$

if $G=⟨a⟩$, $G$ is cyclic and $a$ is its generator