Question

prove first isomorphism theorem

Answer 1

The proof consists of several parts which we will give for completeness. Let $\mathrm{K}$ denote kerf. The following calculation validates that for every $\mathrm{g} \in \mathrm{G}$ and $\mathrm{k} \in \mathrm{K}$ :

$$\begin{array}{l} f \left(g k g ^ {- 1}\right) = f (g) f (k) f (g) ^ {- 1} \quad (f \text {i s a n h o m o m o r p h i s m}) \\ = f (g) 1 _ {H} f (g) ^ {- 1} \quad (\text {d e f i n i t i o n o f} K) \\ = 1 _ {H} \\ \end{array}$$

Hence, $\mathrm{gkg} - 1$ is in $\mathbf{K}$ . Therefore, $\mathbf{K}$ is a normal subgroup of $\mathbf{G}$ and $\mathbf{G} / \mathbf{K}$ is well-defined.

To prove the theorem we will define a map from $\mathrm{G} / \mathrm{K}$ to the image of $\mathrm{f}$ and show that it is a function, a homomorphism and finally an isomorphism.

Let $\theta: \mathrm{G} / \mathrm{K} \longrightarrow \mathrm{Imf}$ be a map that sends the coset $\mathrm{gK}$ to $\mathrm{f(g)}$ .

Since $\theta$ is defined on representatives we need to show that it is well defined. So, let $g1$ and $g2$ be two elements of $G$ that belong to the same coset (i.e. $g1K = g2K$ ). Then, $g1 - 1g2$ is an element of $K$ and therefore $f(g1 - 1g2) = 1$ (because $K$ is the kernel of $G$ ). Now, the rules of homomorphism show that $f(g1) - 1f(g2) = 1$ and that is equivalent to $f(g1) = f(g2)$ which implies the equality $\theta(g1K) = \theta(g2K)$ .

Next we verify that $\theta$ is a homomorphism. Take two cosets $g1K$ and $g2K$ , then:

$$\begin{array}{l} \theta \left(g _ {1} K \cdot g _ {2} K\right) = \theta \left(g _ {1} g _ {2} K\right) \quad (\text {o p e r a t i o n i n} G / K) \\ = f \left(g _ {1} g _ {2}\right) \quad (\text {d e f i n i t i o n o f} \theta) \\ = f \left(g _ {1}\right) f \left(g _ {2}\right) \quad (f \text {i s a n h o m o m o r p h i s m}) \\ = \theta \left(g _ {1} K\right) \theta \left(g _ {2} K\right) \quad (\text {d e f i n i t i o n o f} \theta) \\ \end{array}$$

Finally, we show that $\theta$ is an isomorphism (i.e. a bijection). The kernel of $\theta$ consists of all cosets $\mathrm{gK}$ in $\mathrm{G} / \mathrm{K}$ such that $\mathrm{f(g)} = 1$ but these are exactly the elements $\mathrm{g}$ that belong to $\mathrm{K}$ so only the coset $\mathrm{K}$ is in the kernel of $\theta$ which implies that $\theta$ is an injection. Let $\mathrm{h}$ be an element of $\operatorname{Imf}$ and $\mathrm{g}$ its pre-image. Then, $\theta(\mathrm{gK})$ equals $\mathrm{f(g)}$ thus $\theta(\mathrm{gK}) = \mathrm{h}$ and therefore $\theta$ is surjective.

The theorem is proved. Some version of the theorem also states that the following diagram is commutative:

were $\pi$ is the natural projection that takes $\mathrm{g} \in \mathrm{G}$ to $\mathrm{gK}$ . We will conclude by verifying this. Take $\mathrm{g}$ in $\mathrm{G}$ then, $\theta(\pi(\mathrm{g})) = \theta(\mathrm{gK}) = \mathrm{f}(\mathrm{g})$ as needed.