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Section Day 9
This is an outline of the topics we covered in the ninth day of class. The skeleton notes are in a handout, which can be printed out using the printer icon at the top right of its section of the page for filling in during class. Filled notes for each day will be posted after class to Canvas.
Handout Tuesday 6/16
Objectives: Advanced Learning Outcomes
During our class meeting, we will work on learning the following. Fluency with these is not expected or required before class.
Given a map between groups, determine if it is a homomorphism and if so compute its kernel
State and apply the following mathematical results: G/Z Theorem, Theorem 10.1/10.2 (Properties of Homomorphisms)
State and apply the First Isomorphism Theorem
Theorem 99 . G/Z Theorem.
If
\(G/Z\) is cyclic, then
\(G\) is Abelian.
Proof.
Theorem 100 . Properties of Homomorphisms: Elements.
Let \(\phi:G\to \overline{G}\) be a homomorphism and \(g\in G\text{.}\) Then
\(\displaystyle \phi(e)=\overline{e}\)
\(\phi(g^n)=(\phi(g))^n\) for all
\(n\in \Z\)
If
\(|g|\) is finite, then
\(|\phi(g)|\) divides
\(|g|\)
\(\Ker(\phi)\) is a subgroup of
\(G\)
\(\phi(a)=\phi(b)\) if and only if
\(a\Ker(\phi)=b\Ker(\phi)\)
If
\(\phi(g)=g'\text{,}\) then
\(\phi^{-1}(g')=g\Ker(\phi)\)
Theorem 101 . Properties of Homomorphisms: Subgroups.
Let \(\phi:G \to \overline{G}\) be a homomorphism and \(H\leq G\text{.}\) Then
\(\displaystyle \phi(H)\leq \overline{G}\)
If
\(H\) is cyclic, then
\(\phi(H)\) is cyclic
If
\(H\) is Abelian, then
\(\phi(H)\) is Abelian
If
\(H \normaleq G\text{,}\) then
\(\phi(H)\normaleq \phi(G)\)
If
\(|\Ker(\phi)|=n\text{,}\) then
\(\phi\) is an
\(n\) -to-1 map from
\(G\) onto
\(\phi(G)\)
If
\(|H|=n\text{,}\) then
\(|\phi(H)|\) divides
\(n\)
If
\(\overline{K}\leq \overline{G}\text{,}\) then
\(\phi^{-1}(\overline{K})\leq G\)
If
\(\overline{K}\normaleq \overline{G}\text{,}\) then
\(\phi^{-1}(\overline{K})\normaleq G\)
If
\(\phi\) is onto and
\(\Ker(\phi)=\{e\}\text{,}\) then
\(\phi\) is an isomorphism from
\(G\) to
\(\overline{G}\)
Corollary 102 . Kernels are Normal.
If
\(\phi:G\to \overline{G}\) is a homomorphism, then
\(\Ker(\phi)\normaleq G\text{.}\)
Theorem 103 . First Isomorphism Theorem.
Let \(\phi:G\to \overline{G}\) be a homomorphism. Then the map \(\Psi:G/\Ker(\phi)\to \phi(G)\text{,}\)
\begin{equation*}
\Psi(g\Ker(\phi))=\phi(g)
\end{equation*}
is an isomorphism. In short, \(G/\Ker(\phi)\isom \phi(G)\text{.}\)
