Math4302 Modern Algebra (Lecture 14)

$G=S_3=\{e,\rho,\rho^2,\tau_1,\tau_2,\tau_3\}$ with $H=\{e,\rho, \rho^2\}$, $\tau_1=(12), \tau_2=(23), \tau_3=(13)$.

Number of distinct coset is $|G|/|H|=2$.

The (left and right) cosets are:

$$\tau_1 H=\tau_2 H=\tau_3 H=\{\tau_1,\tau_2,\tau_3\}\\ H=\rho H=\rho^2 H=\{e,\rho,\rho^2\}$$

For this case, left and right cosets are the same (gives the same partition of $G$).

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$H=\{e,\tau\}$

Left cosets:

$$e H=H=\tau_1 H\\ \rho H=\{\tau_3,\rho\}=\tau_3 H\\ \rho^2 H=\{\tau_2,\rho^2\}=\tau_2 H$$

Right cosets:

$$H=H e=H\tau_1\\ H\tau_2=\{\tau_2,\rho\}=H\rho \\ H\tau_3=\{\tau_3,\rho^2\}=H\rho^2$$

Every subgroup of an abelian group is a normal subgroup.

Prove using direct product of cyclic groups.

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If $G$ is finite, and $|H|=\frac{|G|}{2}$, then $H\trianglelefteq G$.

there are exactly two cosets, and one of them must be $H$, then the left coset $G\setminus H$ will always be the same as the right $G\setminus H$.

$A_n\trianglelefteq S_n$

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If $\phi:G\to G'$ is a homomorphism, then $\ker(\phi)\trianglelefteq G$

We will use the equivalent definition of normal subgroup. ($aha^{-1}\in H$ for all $a\in G, h\in H$)

$\phi(aha^{-1})=phi(a)\phi(h)\phi(a)^{-1}=\phi(a)e'\phi(a)^{-1}=e'$, so $aha^{-1}\in \ker(\phi)$

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Consider $\operatorname{GL}(n,\mathbb{R})$ be all the invertible matrices of size $n\times n$

Let $H=\{A\in \operatorname{GL}(n,\mathbb{R})|\det(A)=1\}$.

$H\trianglelefteq \operatorname{GL}(n,\mathbb{R})$

$\phi:\operatorname{GL}(n,\mathbb{R})\to (\mathbb{R}-\{0\},\cdot)$ where $\phi(A)=\det(A)$

Then $H=\ker(\phi)$

We first show that $1\implies 2$.

$aHa^{-1}\subseteq H$:

If $aH=Ha$, for every $h\in H$, $ah=h'a$ for some $h'$, so $aha^{-1}=h'\in H$.

$H\subseteq aHa^{-1}$:

we have $Ha=aH$, so for every $h\in H$, $ha=ah'$ for some $h'$, so $h=ah'a^{-1}\in aHa^{-1}$.

$2\implies 3$: clear

$3\implies 1$:

$aH\subseteq Ha$. for any $h\in H$, $\forall aha^{-1}\in H$, so $aha^{-1}=h'\in H$, so $ah=h'a\in Ha$ so $aH\subseteq Ha$.

$Ha\subseteq aH$: apply previous part to $a^{-1}$., and $a^{-1}H\subseteq Ha^{-1}$, so $\forall h\in H$ $a^{-1}h=h'a^{-1}\in Ha^{-1}$, so $ha=ah'$.