First isomorphism theorem visualisation on cyclic groups such as $C_4$?

Question

I want to demonstrate the first isomorphism theorem on cyclic groups such as $C_4$. I find it hard to see how this map works.

$C_4$ has the cycle $(13)(24)$. The cycle is a composition function so that

$$(f\circ g)(1)=3,$$ $$(f\circ g)(2)=4,$$ $$(f\circ g)(3)=1,$$ and $$(f\circ g)(4)=2.$$

The isomorphism visualisation with arrows may be better explained in terms of other cyclic examples (eg picture here and picture here). I find it hard to understand the isomorphism theorem, (trivia). Perhaps the demostrations on cyclic groups could help to understand the first isomorphism theorem so:

How can you visualise the first isomorphism theorem on cyclic groups?

Answer 1

To do this, you need another group besides $C_4$. Let's choose $C_2$, and what we'll do is send:

$(1\ 2\ 3\ 4) \mapsto (1\ 3)(2\ 4)\\ (1\ 3)(2\ 4) \mapsto e\\ (1\ 4\ 3\ 2) \mapsto (1\ 3)(2\ 4)\\ e \mapsto e$

Alternately, we could send the generator $x$ of $C_4$ to the generator $y$ of $C_2$, effectively sending:

$x^{k \text{ mod }4} \mapsto y^{k \text{ mod }2}$ (this may be easier to verify as a homomorphism).

We can "get rid" of the $x$ and $y$ by considering the additive groups:

$\Bbb Z/4\Bbb Z$ and $\Bbb Z/2\Bbb Z$ and simply sending $[k]_4 \mapsto [k]_2$.

In any case, all of these are "diagrammatically similar", and we find that:

$C_4/\text{ker }f \cong C_2$.

The kernel of $f$ (if we write our two groups as "powers" of their generators $x$ and $y$, respectively) is $\{e,x^2\}$. Let's call this $K$. The elements of $C_4/K$ are:

$K = \{e,x^2\}\\xK = \{x,x^3\}.$

It should be clear that $KK = K$, that $K(xK) = (xK)K = xK$, and that $(xK)^2 = x^2K = K$.

Thus $xK \mapsto f(x) = y$ is an explicit isomorphism the theorem says exists.

Answer 2

I'm not sure what kind of visualization you're looking for here, but there's really not much to the theorem. Suppose we have an arbitrary group of homomorphism $f:G \to H$. We can make it surjective by restricting $H$ to the image of $H$, and we can make it injective by replacing $G$ by $G/\ker f$ (note that a map $f:G/N \to H$ is the same as a map $f:G \to H$ that vanishes on $N$). The induced map $f:G/\ker f \to \operatorname{im} f$ is then an isomorphism.