# $G = \mathbb{Q} / \mathbb{Z}$ surjective map and kernel isomorphism

Let $G = \mathbb{Q} / \mathbb{Z}$, written additively. For all $n > 0$ how come $p_n(x) = nx$ is a surjective homomorphism from $G \rightarrow G$ and how come the kernel of $p_n(x)$ is isomorphic to $\mathbb{Z_{n}}$? I know that once the above are true that we can conclude that $G$ is isomorphic to a proper quotient of itself, but I can not figure out why the above two are true.

• Surjectivity is easy: if $y\in\Bbb Q/\Bbb Z$ has representative $q\in\Bbb Q$, then the coset $x$ containing $\frac{q}{n}$ satisfies $nx=y$. – Wojowu Apr 24 '16 at 20:23
• ah awesome that makes sense, I still am having trouble understanding the kernel part – WheelofSnow Apr 24 '16 at 20:32
• You can compute the kernel explicitly. Which cosets are mapped to "zero" in this quotient group? – mathguy Apr 24 '16 at 20:37
• Put $\alpha_n=\frac{1}{n}$. Notice that $\ker(p_n)=\langle \overline{\alpha_n}\rangle$, and there is an isomorphisms between $\ker(p_n)$ to $\mathbb{Z}_n$ sending $\overline{\alpha_n}$ to $1$. – Darío G Apr 24 '16 at 20:42
• @wore, would it send to 1 or 0? – WheelofSnow Apr 24 '16 at 20:43

An easy proof results from the snake lemma: you have a commutative diagram of short exact sequences: \begin{alignat*}{3}\DeclareMathOperator{\coker}{coker} 0\longrightarrow &\mathbf Z\longrightarrow &&\mathbf Q\longrightarrow &&\mathbf Q/\mathbf Z\longrightarrow 0\\ m_\mathbf Z &{\downarrow} &m_\mathbf Q&\downarrow & &\downarrow m_{\mathbf Q/\mathbf Z}\\ 0\longrightarrow &\mathbf Z\longrightarrow &&\mathbf Q\longrightarrow &&\mathbf Q/\mathbf Z\longrightarrow 0 \end{alignat*} in which the vertical maps are multiplication by $n$. The snake lemma assets we have an exact sequence $$0\to \ker m_\mathbf Z\to \ker m_\mathbf Q\to\ker m_{\mathbf Q/\mathbf Z}\xrightarrow{\delta} \coker m_\mathbf Z\to \coker m_\mathbf Q\to\coker m_{\mathbf Q/\mathbf Z}\to 0$$ Now $\ker m_\mathbf Z= \ker m_\mathbf Q=0$ and $\coker m_\mathbf Q=0$, since we're in a field. So $\coker m_{\mathbf Q/\mathbf Z}=0$, which means multiplication by $n$ in $\mathbf Q/\mathbf Z$ is surjective, and this exact sequence reduces to $$0\to\ker m_{\mathbf Q/\mathbf Z}\xrightarrow{\delta} \coker m_\mathbf Z\to 0$$ and this means $\delta$ is an isomorphism from $\ker m_{\mathbf Q/\mathbf Z}$ onto $\coker m_\mathbf Z=\mathbf Z/n\mathbf Z$.