0
$\begingroup$

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.

Thanks.

$\endgroup$
  • 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
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\psi(Kx) = \phi(x) = y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.