Abstract Algebra, 3e

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, $\varphi^{-1}(E)\unlhd G$.

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 $y=-x+c$.

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 $\forall h\in H\leq G$.