Question about one-to-one correspondence between left coset and image of Homomorphism Let $G$ be a finite group and $K$ be a homomorphism such that $K : G\rightarrow G^*$ and $H=\ker K$.
Then there is a one-to-one correspondence between the number of left cosets of $\ker K$ and the number of elements in $K[G]$.
But, it seems that if there are $a,b\in G$, then there are two distinct left cosets, namely $aH$ and $bH$. 
How can we prove that $aH$ is not equal to $bH$? 
Since $H=\ker K$, then $a\in aH$. Isn't it possible that there may be some $h\in H$ such that $bh=a$? 
 A: $K$ is a weird symbol for a function.
if $K(a)=K(b)$ then $K(ab^{-1})=K(a)K(b)^{-1}=e$ and so $ab^{-1}$ is in the kernel. So $ab^{-1}$ in $H$, since $H$ is normal $ab^{-1}H=H\implies  ab^{-1}HbH=HbH \implies aH=bH$
On the other hand $aH=bH\implies ab^{-1}H=h\implies K(ab^{-1})=e\implies K(a)K(b^{-1})=e\implies K(A)=K(B)$. 
Therefore $K(a)=K(b)\iff aH=bH$
A: The Emperor of Ice Creams' answer is letter-perfect from a mathematical point of view. Since my comment is rather long, I'm writing it is an answer (but it's only an "illustration").
Suppose $G = D_4 = \langle r,s: r^4 = s^2 = e, sr = r^3s\rangle$ (this is the dihedral group of order 8), and that $G' = \langle a,b: a^2 = b^2 = e, ab = ba\rangle$. This is a non-cyclic group of order $4$.
So explicitly $G = \{e,r,r^2,r^3,s,rs,r^2s,r^3s\}$, and $G' = \{e',a,b,ab\}$.
Define: $\phi: G \to G'$ by: $\phi(r^ks^m) = a^kb^m$. Explicitly:
$\phi(e) = e'$
$\phi(r) = a$
$\phi(r^2) = e'$
$\phi(r^3) = a$ (since in $G',\ a^3 = a^2a = e'a = a$)
$\phi(s) = b$
$\phi(rs) = ab$
$\phi(r^2s) = b$
$\phi(r^3s) = ab$.
To see that $\phi$ is a homomorphism, it suffices (why?) to show that:
$\phi(r)^4 = \phi(s)^2 = \phi(e) = e'$ (which is clear), and that:
$\phi(s)\phi(r) = \phi(r)^3\phi(s)$, that is:
$ba = a^3b$ (again this should be clear because $a^3 = a$, and $G'$ is abelian).
Now the kernel of $\phi$ is $N = \{e,r^2\}$. We compute the cosets:
$N = r^2N = \{e,r^2\}$
$rN = r^3N = \{r,r^3\}$ (Note that $r(r^3)^{-1} = r^2$ is indeed in $N$)
$sN = r^2sN = \{s,r^2s\}$ ($sr^2 = (sr)r = (r^3s)r = r^3(sr) = r^3(r^3s) = r^6s = r^2s$).
$rsN = r^3sN = \{rs,r^3s\}$ ($rsr^2 = r(sr^2) = (r(r^2s) = r^3s$).
Note how beautifully this all works out, $\phi$ is constant on every coset:
$\phi(N) = \{e\}$
$\phi(rN) = \{a\}$
$\phi(sN) = \{b\}$
$\phi(rsN) = \{ab\}$
This means we have a bijection (let's call it $[\phi]: G/N \to G'$). I leave it to you, to PROVE $[\phi]$ is a homomorphism (and thus an isomorphism), the crucial step is noting that:
$[\phi](rsN) = ab = [\phi](rN)[\phi](sN)$.
Note well that even though $s \neq r^2s \in G$, we still have $sN = r^2sN$. This is because cosets "clump" elements of $G$ together. When we say $sN$, the COSET of $N$ we mean is clear, but although $s$ "identifies" it, $r^2s$ would do just as well. And as far as $[\phi]$ is concerned, it cannot tell them apart (And why is this? Because $\phi$ sends both to the same target).
What is going on, here, geometrically? Algebraically, we can see we've "collapsed" the subgroup $N$ to a single element (our new identity). Since $N$ has two elements, this halved the size of our group. Now $D_4$ is the symmetry group of the square, and the geometric consequence of this, is that we have "identified" two opposite ends of a diagonal, collapsing our square down to a line segment (sometimes called a "di-gon").
