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While studying mathematics I have often asked myself the question, "Who on earth thought about all these equations, theorems, definitions, and proofs?". If you have ever found yourself wondering the same thing, or something similar, then you are in the right place. The purpose of this blog is to shed some light on the faces behind great mathematics.
~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:
Injective:
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.