The First Isomorphism Theorem Proof

~A proof for my Dad, with help from Ben, my teacher.

Definitions

Isomorphism: 

* To be an isomorphism the function must be a homomorphism and be surjective (everyone has a room) and injective (no one is sharing a room), this means it is a bijection.

Homomorphism:

We say that a function f from G to H is a group homomorphism if f preserves the group structures, which means $f(a\cdot b)=f(a)\times f(b)$ for every a and b in G. 

Surjective:

• A function f from A to Bf is surjective if and only if (y∈B) implies (there exists x∈A such that f(x)=y).

Injective:

• A function f from A to Bff is injective if and only if (x1,x2∈A such that f(x1)=f(x2)) implies (x1=x2).

The First Isomorphism Theorem that I prove below:

Let f from G to H be a surjective homomorphism with kernel $K$. Because we know that $f(x)=f(y)$ for any $y\in Kx$ (elements in the same coset of the kernel have the same image under $f$), then we can define a map $\phi:G/K\to H$ by defining $\phi(Kg)=f(g)$. This map $\phi$ is always an isomorphism.

POOF TIME!!!

To prove this theorem we must show that $\phi: G/K\to H$ is injective and surjective, thus it is an isomorphism.

First, let $f:G\to H$ be a surjective homomorphism with kernel $K$. Also, let $\phi:G/K\to H$ be defined for all $gK\in G/K$ by $\phi(gK)=f(g)$.

Now let's show that $\phi$ is a function. 

This means we must show that for every $X\in G/K$ there exists a unique $h\in H$ such that $\phi(X)=h$.

To start let $X\in G/K$.

Next pick $g, j\in G$ such that $X=gK=jK$. 

Since there are different ways to represent the same coset, we have chosen two of them, and will now show that the choice $g$ or $j$ does not change the outcome of the rules given by $\phi$. (Similar to how some people know me as Haylee Zierow and some know me as Haylee Lau, but despite the different names I am the same person).

We will now show that $\phi(X)\in H$ and that $\phi(X)$ is unique.

We compute $$\phi(X)=\phi(gK)=f(g)\in H.$$

Also, $$\phi(X)=\phi(jK)=f(j)\in H.$$

This shows that $\phi(X)\in H$.

Now, since $gK=jK$, we know that $g\in jK$ (refer to problem $82$).

This implies that $g=jk$ for some $k\in K$ which implies that $$f(g)=f(jk)=f(j)f(k)=f(j)e_H=f(j).$$

Since $f(g)=f(j)$ it follows that $\phi(gK)=\phi(jK)$ because of how $\phi$ is defined, hence $\phi(X)$ is unique.

We have now shown that $\phi$ is a function. 

Next, with $gK, jK\in G/K$ we will show that $\phi(gKjK)=\phi(gK)\phi(jK)$.

We compute $$\phi(gKjK)=\phi((gj)K)=f(gj)=f(g)f(j)=\phi(gK)\phi(jK).$$  

So $\phi$ is a group homomorphism.

Now suppose that $\phi(gK)=\phi(jK)$ for some $gK, jK\in G/K$.

Then we know that $f(g)=f(j)$ and we also know that $f(g)(f(j))^{-1}=f(gj^{-1})=e_H$. 

This shows that $gj^{-1}\in K$, so $gK=jK$, thus we know $\phi$ is injective.

Now suppose that $h\in H$.

Since $f$ is surjective, that means there exists $g\in G$ such that $f(g)=h$.

We then have $$\phi(gK)=f(g)=h.$$

Thus $\phi$ is surjective.

Therefore, $\phi:G/K\to H$ is an isomorphism.