# First Isomorphism Theorem: Let $\phi : G\rightarrow G'$ be a homomorphism of groups. Then $G/\ker(\phi ) \cong \phi (G)$.

I was looking at a proof to this theorem and I found one, but couldn't understand some parts of the proof...The proof goes something like this:
In the part above, I couldn't understand how: "$$a = a'\circ g$$", and what exactly $$a'$$ is?!
After this the proof carries on as...
..again I am confused as to how "$$\ker \psi = K \Rightarrow \rm Injectivity$$"
. Then the proof goes on as...
...again I have no idea how.

Thanks.

• $a\in Kg\implies a=a'g$ for some $a'\in K$. – Thomas Shelby Jun 27 at 5:17
• $\psi$ will be injective iff its kernel consists just of the identity element; the identity element of $G/K$ is the trivial coset $K$. – Lord Shark the Unknown Jun 27 at 5:20

Well, everything is there in the proof.

In view of well-definedness, one has to check that the whole coset

$$Kg = \{ag\mid a\in K\}$$

maps to $$\phi(g)$$. Here $$g$$ is a representative and every other representative is of the form $$ag$$ for some $$a\in K$$.

For each $$a\in K$$, we have $$Kg = (Ka)g = K(ag)$$.

Then $$\psi(K(ag)) = \phi(ag)=\phi(a)\phi(g)=\phi(g)$$, since $$\phi(a)=1$$ (unit element in $$G$$).

So no matter how one chooses the representative of the coset it maps to the same element.

The first part of the proof shows that for any two elements, i.e., cosets $$A$$ and $$B$$ in $$G/K$$ if $$A$$ and $$B$$ are equal then $$\psi(A)$$ and $$\psi(B)$$ are also equal, which means that $$\phi$$ is actually a function and we say that $$\phi$$ is well-defined in that case. Since any coset $$X$$ in $$G/K$$ is the set $$Kx$$ of all products $$kx$$ where $$x$$ is some element in $$G$$ and $$k$$ runs through all elements in $$K$$, $$\psi$$ is well-defined if and only if for any two elements $$a$$ and $$b$$ in $$G$$ if $$Ka$$ and $$Kb$$ are equal then $$\phi(a)$$ and $$\phi(b)$$ are also equal. Now, for any two elements $$a$$ and $$b$$ in $$G$$, $$Ka = Kb$$ if and only if $$a \in Kb$$ ( equivalently $$b \in Ka$$ ) since the cosets $$Ka$$ and $$Kb$$ are equivalence classes of a equivalence relation. And $$a \in Kb$$ means that $$a = kb$$ for some $$k$$ in $$K$$. Hence, $$\phi$$ is well-defined if and only if for any two elements $$a$$ and $$b$$ in $$G$$ if $$a = kb$$ for some element $$k$$ in $$K$$ then $$\phi(a) = \phi(b)$$. The latter is true for $$\phi(a) = \phi(kb) = \phi(k)\phi(b) = \phi(b)$$ ( notice that $$\phi(k)$$ is $$1$$ because $$k$$ is an element of the kernel $$K$$ of $$\phi$$ ), and so, the former is true.

In the second part, the author made a careless mistake or had promised to use an abuse of language. Instead of ker $$\psi = K$$, the author should have written ker $$\psi = \{ K \}$$. With this correction, ker $$\psi = \{ K \}$$ implies that, for any two elements $$a$$ and $$b$$ in $$G$$ such that $$\psi(Ka) = \psi(Kb)$$, $$Kab^{-1} = K$$ which is equivalent to $$Ka = Kb$$.

Finally, the proof ends with saying that every element $$y$$ in $$\phi(G)$$ is of form $$\phi(x)$$ for some $$x$$ in $$G$$ by definition and so $$\psi(Kx) = \phi(x) = y$$.