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109 lines
3.6 KiB
Markdown
109 lines
3.6 KiB
Markdown
# Math4302 Modern Algebra (Lecture 17)
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## Subgroup
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### Normal subgroup
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#### Fundamental theorem of homomorphism
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If $\phi: G\to G'$ is a homomorphism, then the map $f:G/\ker(\phi)\to\phi(G)$ given by $f(a\ker(\phi))= \phi(a)$ is well defined and is an isomorphism.
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$\ker(\phi)\trianglelefteq G$, and $\phi(G)=\{\phi(a)|a\in G\}\leq G'$
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<details>
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<summary>Proof</summary>
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First we will prove the well definedness and injectivity of $f$.
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We need to check the map will not map the same coset represented in different ways to different elements.
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Suppose $a\ker(\phi)=a'\ker(\phi)$, then $a^{-1}b\in \ker(\phi)$, this implies $\phi(a^{-1}b)=e'$ so $\phi(a)=\phi(b)$.
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Reverse the direction to prove the converse.
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The injective property is trivial.
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Next, we will show that $f$ is a homomorphism.
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$$
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\begin{aligned}
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f(a\ker \phi b\ker \phi)&=f(ab\ker \phi)\\
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&=\phi(ab)\\
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&=\phi(a)\phi(b)\text{ since $\phi$ is homomorphism}\\
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&=f(a\ker \phi)f(b\ker \phi)\\
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\end{aligned}
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$$
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We also need to show $f$ is surjective:
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If $\phi(a)\in \phi(G)$, then $f(a\ker \phi)=\phi(a)$
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</details>
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<details>
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<summary> Examples for application of theorem</summary>
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If $\phi$ is injective, then $\ker \phi=\{e\}$, so $G\simeq \phi(G)\leq G'$
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---
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If $\phi$ is surjective, then $\phi(G)=G'$, so $G/\ker\phi \simeq G'$
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---
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Let $\phi:G\to G'$ be any homomorphism between two groups.
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Then there exists a surjective map that $G\to G/\ker(\phi)$ by $a\mapsto aN$.
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And there exists a injective map that $G/\ker(\phi)\to G'$ by $a\ker(\phi)\mapsto \phi(a)$
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</details>
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> [!TIP]
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>
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> In general, if $N\trianglelefteq G$, then we have a homomorphism $\phi:G\to G/N$ where $a\mapsto aN$. $\ker(\phi)=N$
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If $\phi:G\to G'$ is a non-trivial homomorphism, and $|G|=18$ and $|G'|=15$, then what is $|\ker\phi|$?
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<details>
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<summary>Solution</summary>
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Note that $G/\ker(\phi)\simeq \phi(G)$ since $|G'|=15$, then $|G/\ker(\phi)|=1,3,5$ or $15$. But $|G/\ker(\phi)|=\frac{|G|}{|G/\ker(\phi)|}=\frac{18}{1,3,5,15}$ so $|G/\ker(\phi)|=1$ or $3$.
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Since $\phi$ is not trivial, $G\neq \ker(\phi)$, so $|G/\ker(\phi)|=3$
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So $|\ker(\phi)|=6$.
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Example: $\mathbb{Z}_{18}\to \mathbb{Z}_3\times \mathbb{Z}_5$ by $[a]\mapsto ([a\mod 3],[0])$. $\ker(\phi)=\{0,3,6,9,12,15\}$
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</details>
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What is $\mathbb{Z}_{12}/\langle 4\rangle$? \langle 4\rangle=\{0,4,8,\}$
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<details>
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<summary>Solution</summary>
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The quotient of every cyclic group is also cyclic.
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Let $G=\langle a\rangle $ and $N\trianglelefteq G$, then $G/N=\langle aN\rangle$ if $bN\in G/N$, then $b=a^k$ for some $k$. So $bN=a^k N=(aN)^k$
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So $\mathbb{Z}_{12}/\langle 4\rangle$ is a cyclic group of order $4$, so it is isomorphic to $\mathbb{Z}_4$.
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</details>
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What is $\mathbb{Z}\times \mathbb{Z}/\langle (1,1)\rangle$? Let $N=\langle (1,1)\rangle$.
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<details>
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<summary>Solution</summary>
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This is isomorphic to $\mathbb{Z}$, sending the addition over one axis $\mathbb{Z}$. Show the kernel is $N$
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</details>
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#### There is only two group of order 4
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Every group of order $4$ is isomorphic to either $\mathbb{Z}_4$ or $(\mathbb{Z}_2\times \mathbb{Z}_2,+)$
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If $|G|=4$ and there is an element of order $4$ in $G$. then $G$ is cyclic, so $G\simeq \mathbb{Z}_4$.
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Otherwise since $|\langle a\rangle|||G|=4$ every $a\neq G$. Let $G=\{e,a,b,c\}$, Then $a^2=b^2=c^2=e$, and $ab=c (ab\neq e,ab\neq a,ab\neq b)$, so $G$ is isomorphic to $(\mathbb{Z}_2\times \mathbb{Z}_2,+)$
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So we can use this property to define $G\to \mathbb{Z}_2\times \mathbb{Z}_2$ by $\phi(e)=(0,0), \phi(a)=(0,1), \phi(b)=(1,0), \phi(c)=(1,1)$ order of $a,b,c$ does not matter.
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