Let $G$ be a multiplicative group of non-zero complex numbers. Consider the group homomorphism from $\phi:G→G$ given by $\phi(z)=z^4$
a) identify $H$, the kernel of $\phi$
b) identify (upto isomorphism) the quotient space $G/H$
My attempt:
a) $H=\ker\phi=\{z\in G:\phi(z)=1\}$
i.e $\phi(z)=z^4=1$
i.e $(\cos x + i \sin x)^4=1$
i.e $\cos4x + i \sin4x=1$
i.e $\cos4x=1,\sin4x=0$
solving above equation we will get $z=e^{iπn/2}$
so $H=\{e^{in\pi/2},n\in \Bbb{N}\}$
Best Answer
I suppose that $G=(\mathbb{C}\setminus\{0\},\times)$. Then $\ker\varphi$ is, as you wrote, the set of those $z\in\mathbb{C}\setminus\{0\}$ such that $z^4=1$. Therefore$$\ker\varphi=\{1,-1,i,-i\}.$$
The quotient space can be identified with $\mathbb{C}\setminus\{0\}$ itself. That's because the map$$\begin{array}{ccc}G&\longrightarrow&G\\z&\mapsto&z^4\end{array}$$is a surjective group homomorphism, and therefore it induces a isomorphism between $G/H$ and $G$.