====== Abstract Algebra, 3e ======
A thorough introduction to abstract algebra, from groups to rings to modules to vector spaces and so on.

===== Notes =====
==== Quotient Groups and Homomorphisms ====
I've always had a tenuous grasp on cosets and quotient groups, but this book is helping me make sense of them. Unlike previous texts which use cosets to define quotient groups, this one uses the structure of quotient groups to describe cosets. Its initial focus is on how group homomorphisms give rise to quotient groups. That is, the fibers of a homomorphism $\varphi:G\to H$ with kernel $K$ naturally form the group $G/K$. It reduces cosets to objects (in this case subsets of a group) that may or may not possess a group structure, then lays out the conditions under which those objects do form a group (the cosets must be of a normal subgroup). Similarly, normal subgroups are not defined in terms of cosets, but rather in terms of quotient groups. 

<WRAP center round help 80%>
How are quotient groups analogous to integer quotients? Are the two even analogous?
</WRAP>

++++ Exercise 3.1.1 |
> Let $\varphi:G\to H$ e a homomorphism and let $E\leq H$. Prove that $\varphi^{-1}(E)\leq G$. If $E\unlhd H$, prove that $\varphi^{-1}(E)\unlhd G$. Deduce that $\ker \varphi\unlhd G$.

Since $E\leq H$, $1_H\in E$, so $1_G\in\varphi^{-1}(E)$ (that is, $\varphi^{-1}(E)\neq\emptyset$). Let $x,y\in\varphi^{-1}(E)\subseteq G$. $\exists a,b\in E$ s.t. $\varphi(x)=a, \varphi(y)=b$. $\varphi(xy^{-1})=\varphi(x)\varphi(y)^{-1}=ab^{-1}$. Since $E\leq H$, $ab^{-1}=\varphi(xy^{-1})\in E$, so $xy^{-1}\in \varphi^{-1}(E)$. By the subgroup criterion, $\varphi^{-1}(E)\leq G$, as desired.

Now suppose $E\unlhd H$. Let $x\in\varphi^{-1}(E), g\in G$. $\exists h\in H$ s.t. $\varphi(g)=h$. $\varphi(gxg^{-1})=h\varphi(x)h^{-1}$. Since $\varphi(x)\in E$ and $E\unlhd H$, $h\varphi(x)h^{-1}\in E\implies g\varphi^{-1}(E)g^{-1}\subseteq\varphi^{-1}(E)$. By theorem 6, $\varphi^{-1}(E)\unlhd G$. $\square$
++++

++++ Exercise 3.1.7 |
To see that $\pi$ is a homomorphism, let $(x_1, y_1), (x_2, y_2)\in \mathbb{R}^2$. $\pi((x_1,y_1)+(x_2,y_2)) = \pi((x_1+x_2,y1+y2)) = (x_1+x_2)+(y_1+y_2) = (x_1+y_1)+(x_2+y_2) = \pi((x_1, y_1))+\pi((x_2,y_2))$.

To how that $\pi$ is surjective, let $a\in\mathbb{R}$. $a=a+0$, so $\pi((a, 0)=a$. (As an aside, it is immediately clear that $\pi$ is **not** injective, since $\pi((0,a))$ also equals $a$.)

$\ker\pi$ is the line $y=-x$. For any $c\in\mathbb{R}$, the fiber above $c$ is the line $y=-x+c$. $\square$
++++

++++ Exercise 3.1.24 |
> Prove that if $N\unlhd G$ and $H$ is any subgroup of $G$, then $(N\cap H)\unlhd H$.

First we show that $(N\cap H)\leq H$. Note that $1\in (N\cap H)$ since $N,H\leq G$; that is, $(N\cap H)\neq\emptyset$. Let $x,y\in(N\cap H)$. $N,H\leq G\implies xy^{-1}\in N,H\implies xy^{-1}\in(N\cap H)$. Hence $(N\cap H)\leq H$. 

Next we show $(N\cap H)\unlhd H$. By Theorem 3.6, it suffices to show that $h(N\cap H)=(N\cap H)h~\forall h\in H$ (i.e., $hn=nh~\forall h\in H, n\in(N\cap H)$). Since this already holds $\forall g\in G$ (and since $(N\cap H)\leq H$), it must hold $\forall h\in H\leq G$. $\square$
++++

++++ Exercise 3.2.22 |
> Use Lagrange's Theorem in the multiplicative group $(\mathbb{Z}/n\mathbb{Z})^\times$ to prove Euler's Theorem: $a^{\varphi(n)}\equiv 1\mod n$ for every integer $a$ relatively prime to $n$.

Let $a\in (\mathbb{Z}/n\mathbb{Z})^\times$. Note that $\gcd(n,a)=1$ and $|(\mathbb{Z}/n\mathbb{Z})^\times|=\varphi(n)$ by definition of $(\mathbb{Z}/n\mathbb{Z})^\times$. By corollary 9, $a^{|(\mathbb{Z}/n\mathbb{Z})^\times|}=1$. By analogy (be more specific), $a^{|\varphi(n)|}\equiv 1\mod n$. $\square$
++++

Corollary 17 closely resembles some important results from linear algebra (see [[Linear Algebra Done Right]], 3.16 and 3.22). This makes sense, as linear transformations are definitionally group homomorphisms. 

<WRAP center round box 60%>
**Proof of Corollary 17(1)**

($\implies$) Suppose $\varphi$ is injective; we must show that $\ker\varphi = \{1\}$. $\varphi(1)=1\implies 1\in\ker\varphi\implies\{1\}\subseteq\ker\varphi$. Let $g\in \ker\varphi$. $\varphi(g)=1=\varphi(1)$; since $\varphi$ is injective, we must have $g=1$ and thus $\ker\varphi=\{1\}$.

($\impliedby$) Now suppose $\ker\varphi=\{1\}$; we must show that $\varphi$ is injective. Let $a,b\in G$ such that $\varphi(a)=\varphi(b)$. Then $\varphi(a)*\varphi(b)^{-1}=1\iff\varphi(a*b^{-1})=1\implies a*b^{-1}\in\ker\varphi$. By hypothesis, it must be the case that $a*b^{-1}=1$; that is, $a=b$. Hence $\varphi$ is injective, as desired.
</WRAP>

<WRAP center round todo 80%>
Revisit normalizers, centralizers, and stabilizers (§2.2, p. 49) and take notes on this page. They have appeared more often than expected when I skimmed past them.
</WRAP>

{{tag>math algebra}}