G2 - normal subgroup test

if normal, for ant $x\in G$ and $h\in H$, there is an $h^{\prime }\in H$ such that $xh=h^{\prime }x$

thus $xh{x}^{-1}=h^{\prime }$, and therefore, $xH{x}^{-1}\subseteq H$

conversely, $xH{x}^{-1}\subseteq H⟹aH{a}^{-1}\subseteq H⟹aH\subseteq Ha$, and $a^{-1}H(a^{-1}{)}^{-1}=a^{-1}Ha\subseteq H⟹Ha\subseteq aH$

every subgroup of an abelian group is normal

the center of a group $Z(G)$ is always normal

the alternating group $A_n$ of even permutations is a normal subgroup of $S_n$, eg: for $(12)\in S_n$ and $(123)\in A_n$, $(12)(123)\ne (123)(12)$ but $(12)(123)=(132)(12)$ and $(132)\in A_n$

every subgroup of $D_n$ consisting only of rotations is normal in $D_n$

let $H⊲G$ and $K\le G$, then $HK=\{hk\mid h\in H,k\in K\}\le G$

in $HK$, $e=ee$

for any $a=h_1{k}_1$ and $b=h_2{k}_2$, there exists an $h^{\prime }\in H$ such that $a{b}^{-1}=h_1{k}_1{k}_2^{-1}{h}_2^{-1}=(h_1{h}^{\prime })(k_1{k}_2^{-1})$, so $a{b}^{-1}\in HK$

if a group $G$ has a unique subgroup $H$ of some finite order, then $H$ is normal in $G$

for any $g\in G$, the inner automorphism ${\varphi }_g:H\to gH{g}^{-1}$, so $gH{g}^{-1}\le G$ and $|gH{g}^{-1}|=|H|$

$H=SL(2,\mathbf{R})⊲GL(2,\mathbf{R})=G$

let $x\in G$ and $h\in H$, then $\text{det}(xh{x}^{-1})=(\text{det }x)(\text{det }h)(\text{det }x{)}^{-1}=(\text{det }x)(\text{det }x{)}^{-1}=1$

thus, $xh{x}^{-1}\in H$, and therefore, $xH{x}^{-1}\subseteq H$