Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.19
Let $G = M = \langle u,v \ |\ u^2 = v^8 = 1, vu = uv^5 \rangle$ and let $H = \langle v^4 \rangle = \{ 1, v^4 \}$.
(1) Show that the order of $G/H$ is 8.
(2) Exhibit each element of $G/H$ in the form $\overline{u}^a\overline{v}^b$ for some integers $a$ and $b$.
(3) Find the order of each element of $G/H$.
(4) Write each of the following elements in the form $\overline{u}^a \overline{b}$: $\overline{vu}$, $\overline{uv^{-2}u}$, and $\overline{u^{-1}v^{-1}uv}$.
(5) Prove that $G/H$ is isomorphic to $Z_2 \times Z_4$.
Solution:
(1)&(2) We can find the elements of $G/H$ explicitly: $$G/H = \{ \overline{1} = \{ 1, v^4 \}, \overline{v} = \{ v,v^5 \}, \overline{v}^2 = \{ v^2,v^6 \}, \overline{v}^3 = \{ v^3, v^7 \}, \overline{u} = \{ u, uv^4 \}, \overline{u}\,\overline{v} = \{ uv,uv^5 \}, \overline{u}\,\overline{v}^2 = \{ uv^2,uv^6 \}, \overline{u}\,\overline{v}^3 = \{ uv^3,uv^7 \} \}.$$(3)
| —-$x$—- | ————————Reasoning———————— | Order |
| $\overline{1}$ | 1 | |
| $\overline{v}$ | 4 | |
| $\overline{v}^2$ | $ (\overline{v}^2)^2 = \overline{v^4} = \overline{1}$ | 2 |
| $\overline{v}^3$ | $ |x^{-1}| = |x|$ | 4 |
| $\overline{u}$ | 2 | |
| $\overline{u}\,\overline{v}$ | $(\overline{u}\,\overline{v})^2 = \overline{uvuv} = \overline{u^2 v^6} = \overline{v}^2$, $\overline{u}\,\overline{v}\overline{v}^2 = \overline{u}\,\overline{v}^3$, and $\overline{u}\,\overline{v}\,\overline{u}\,\overline{v}^3 \overline{uvuv^3} = \overline{u^2v^8} = \overline{1}$. | 4 |
| $\overline{u}\,\overline{v}^2$ | $ (\overline{u}\,\overline{v}^2)^2 = \overline{uv^2uv^2} = \overline{u^2v^{12}} = \overline{v^4} = \overline{1}$ | 2 |
| $\overline{u}\,\overline{v}^3$ | $|x| = |x^{-1}|$ | 4 |
(4) We have $$\overline{vu} = \overline{uv^5} = \overline{u}\,\overline{v},$$ $$\overline{uv^{-2}u} = \overline{uv^6u} = \overline{u^2v^{30}} = \overline{v^6} = \overline{v}^2,$$ and $$\overline{u^{-1}v^{-1}uv} = \overline{uv^7uv} = \overline{u^2v^{36}} = \overline{v^4} = \overline{1}.$$ (5) We let $Z_2 = \langle x \rangle$ and $Z_4 = \langle y \rangle$. Define a mapping $\varphi : Z_2 \times Z_4 \rightarrow G/H$ by $(x^a,y^b) \mapsto \overline{u}^a\overline{v}^b$.
(Homomorphism) We have $$\varphi((x^a,y^b)(x^c,y^d)) = \varphi(x^{a+c},y^{b+d}) = \overline{u}^{a+c}\overline{v}^{b+d} = \overline{u}^a\,\overline{u}^c\,\overline{v}^b\,\overline{v}^d.$$ Note that $\overline{v}\,\overline{u} = \overline{vu} = \overline{uv^5} = \overline{u}\,\overline{v}$, so we have $$\overline{u}^a\,\overline{u}^c\,\overline{v}^b\,\overline{v}^d = \overline{u}^a\,\overline{v}^b\,\overline{u}^c\,\overline{v}^d= \varphi(x^a,y^b)\varphi(x^c,y^d).$$ Thus $\varphi$ is a homomorphism.
(Surjective) We saw above that every element of $G/H$ is of the form $\overline{u}^a\overline{v}^b$ for some $0 \leq a < 2$ and $0 \leq b < 4$; clearly, then, $\varphi$ is surjective. Now we know that $|G/H| = |Z_2 \times Z_4| < \infty$, so that $\varphi$ is bijective. Thus $G/H \cong Z_2 \times Z_4$.
