TITLE: Intuitive Explanation of Morphism Theorem
QUESTION [5 upvotes]: Is there an intuitive explanation for the morphism theorem from introductory abstract algebra?

First Morphism Theorem: Let $K$ be the kernel of the group morphism $f: G \to H$.  Then $G/K$ is isomorphic to $\text{Im } f$, and the isomorphism $\psi : G/K \to \text{Im } f$ is defined by $\psi(Kg) = f(g)$.

I follow the steps of the proof of the theorem, but find the theorem very nonintuitive to conceptualize.

REPLY [2 votes]: Maybe look at some simple examples.
Consider the groups $G = (\mathbb{Z},+)$  and $H = (\mathbb{C}^*, \times)$. Define $f : G \to H$ by $f(n) = (-1)^n$ for $n \in \mathbb{Z}$. You can check that $f$ is a group morphism. We can see that $\textrm{Im}(f) = \{-1,1\}$ and $\ker(f) = 2\mathbb{Z}$. In that case, the theorem states that

The value of $f(k) = (-1)^k$ only depends the class of $k$ modulo $\ker(f) = 2\mathbb{Z}$ (i.e. the parity of $k$) : it's $-1$ if $k$ is odd, and $1$ if $k$ is even. 
The previous point allows you to define $\psi : \mathbb{Z}/2\mathbb{Z} \to \{-1,1\}$ by $\psi(k + 2 \mathbb{Z}) = (-1)^k$ and this is a group isomorphism.

For another (similar) example, fix $n \in \mathbb{N}^*$ and define $f : G \to H$ by $f(k) = e^{\frac{2ik\pi}{n}}$. Denote $\mathbb{U}_n = \left\{1, e^{\frac{2i\pi}{n}}, e^{\frac{2i2\pi}{n}}, \ldots,e^{\frac{2i(n-1)\pi}{n}}\right\} = \{z \in \mathbb{C}, z^n = 1\}$. Then $\textrm{Im}(f) = \mathbb{U}_n$, $\ker(f) = n \mathbb{Z}$ and the theorem yields an isomorphism $\psi : \mathbb{Z}/n\mathbb{Z} \to \mathbb{U}_n$ defined by $\psi(k+n\mathbb{Z}) = e^{\frac{2ik\pi}{n}}$ (as in the case $n = 2$, the value $e^{\frac{2ik\pi}{n}}$ only depends on the class of $k$ modulo $n$, so $\psi$ is well defined).
In a similar fashion, consider $G = (\mathbb{R},+)$, $H = (\mathbb{C}^*, \times)$ and $f: x \mapsto e^{i x}$. We have $\text{Im}(f) = \{z \in \mathbb{C}, |z| = 1\}$ and $\ker(f) = 2\pi\mathbb{Z}$. In that case the theorem gives you an isomorphism $\psi : \mathbb{R}/2\pi\mathbb{Z} \to \{z \in \mathbb{C}, |z| = 1\}$ given by $\psi(x + 2\pi\mathbb{Z}) = e^{i x}$ (once again this is well defined because $e^{ix}$ only depends on "$x$ modulo $2\pi$"). As a matter of fact, the reciprocal of $\psi$ in that case is what we usually call the argument (the argument of a complex number is only defined "up to a multiple of $2\pi$", which essentially means it really is an element of $\mathbb{R}/2\pi\mathbb{Z}$).
For yet another example, consider $G = H = (\mathbb{C}^*, \times)$. Fix $n \in \mathbb{Z}$ with $n \ne 0$ and define $f : G \to H$ by $f(z) = z^n$. Then $f$ is a morphism and $\textrm{Im}(f) = \mathbb{C}^*$ (the equation $x^n = y$ always has a solution because $\mathbb{C}$ is algebraically closed) and $\ker(f) = \mathbb{U}_n$. The theorem gives you an isomorphism $\psi : \mathbb{C}^*/\mathbb{U}_n \to \mathbb{C}^*$ defined by $\psi(z\mathbb{U}_n) = z^n$. You could think of of $\psi^{-1}(z)$ as $\sqrt[n]{z}$ but because there are $n$ possible choices for an $n^{\textrm{th}}$ root of $z$ (all of which only differ by an element of $\mathbb{U}_n$), it is not a well defined number, but a class modulo $\mathbb{U}_n$.