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:
part-1 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...
part-2 ..again I am confused as to how "$\ker \psi = K \Rightarrow \rm Injectivity$"
. Then the proof goes on as...
part-3 ...again I have no idea how.


  • 1
    $\begingroup$ $a\in Kg\implies a=a'g$ for some $a'\in K$. $\endgroup$ – Thomas Shelby Jun 27 at 5:17
  • 2
    $\begingroup$ $\psi$ will be injective iff its kernel consists just of the identity element; the identity element of $G/K$ is the trivial coset $K$. $\endgroup$ – 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 $\psi(Kx) = \phi(x) = y$.


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