Math4302 Modern Algebra (Lecture 15)

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

For example, if $\det:GL(n,\mathbb{R})\to \mathbb{R}-\{0\}$ is a homomorphism, then

$$H=\ker(\det)=\{A\in GL(n,\mathbb{R})|\det(A)=0\}=SL(n,\mathbb{R})\trianglelefteq GL(n,\mathbb{R})$$

First, suppose $H\trianglelefteq G$, and $aH=a'H$m and $bH=b'H$, we want to show that $abH=ab'H$.

It is enough to show that $(ab)^{-1}a'b'=b^{-1}a^{-1}a'b'\in H$.

$aH=a'H\implies a^{-1}a'\in H$, and $bH=b'H\implies b^{-1}b'\in H$. Note that by proposition of normal group, $gHg^{-1}\subseteq H$ for any $g\in G$, so let $g=b^{-1}$, $h=a^{-1}a$.

Therefore $b^{-1}(a^{-1}a')(b^{-1})^{-1}=b^{-1}a^{-1}a'b\in H$, since $b^{-1} b'\in H$, then $b^{-1}a^{-1}a'b'\in H$.

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Conversely, suppose this operation is well defined, then we show that $ghg^{-1}\in H$ for any $g\in G, h\in H$.

Note that $hH=eH$, the well-defineness implies that $(hH)(g^{-1}H)=(eH)(g^{-1}H)=g^{-1}H$. So $ghg^{-1}\in H$. (add $g$ on the left)

$aH=bH\iff a^{-1}b\in H$, or equivalently $aH=bH\iff b^{-1}a\in H$.

This operation is well defined by condition above.

• Identity: $eH=H$
• Inverse: $(aH)^{-1}=a^{-1}H$
• Associativity: $(aH bH)cH=aH(bH cH)=abcH$

Recall from previous lectures, $G=S_3$ with $H=\{e,\tau_1\}$, with $\tau_1=(12), \tau_2=(23), \tau_3=(13)$.

• $\{e,\tau_1\}=\tau_1 H=H$
• $\{\tau_2,\rho_2\}=\tau_2 H=\rho_2 H$
• $\{\tau_3,\rho\}=\tau_3 H=\rho H$

And $(\tau_2 H)(\tau_3 H)=\tau_2 \tau_3 H=\rho H$.

However, if we take $\rho^2\in \tau_2 H$, and $\rho\in \tau_3 H$, $\rho^2\rho =e$. This is not in $\rho H$.

This is not well defined since $H$ is not normal.

$5\mathbb{Z}\trianglelefteq \mathbb{Z}$, the cosets are $5\mathbb{Z}, 1+5\mathbb{Z}, 2+5\mathbb{Z}, 3+5\mathbb{Z}, 4+5\mathbb{Z}$.

Here $5\mathbb{Z}$ is the identity in the factor group.

And $\mathbb{Z}/5\mathbb{Z}\simeq \mathbb{Z}_5$