Why is this map well-defined?

Question

Let $G$ a finite group and $H$ and $K$ two sub-groups of $G$.

Why is the map
\begin{array}{rcl} \Psi: G/({H\cap K}) & \longrightarrow &G/H \times G/K \\ g(H\cap K) & \longmapsto & (gH, \ gK) \\ \end{array} well-defined ?

I do not understand why the fact of having $$g(H\cap K)=g'(H\cap K) \implies (gH, \ gK)=(g'H, \ g'K)$$ answers the question.

Concretely what does that kind of question mean?

NB : It's an important condition to prove Poincaré's Lemma

Answer 1

The elements of a quotient group $G/N$ are cosets $$ gN = \{ gn : n\in N \}. $$ Different $g$ might yield the same cosets of $N$. In fact $gN=g'N$ if and only if $g^{-1}g'\in N$: Let $g^{-1}g'\in N$, take $g'n\in g'N$, then $$g'n = g\underbrace{g^{-1}g'n}_{\in N} \in g'N,$$ so $g'N\subseteq gN$. Using the same trick we show that $gN\subseteq g'N$, so $gN=g'N$. Conversely, having $gN=g'N$, we have $g'=g'e\in g'N=gN$, so $g'=gn$ for some $n\in N$. Then $g^{-1}g'=n\in N$.

Now whenever you define a map $f:G/N\to X$ using a definition of the form $gN\mapsto \ldots$, you have to check that taking different representatives of $gN$ yields the same image, i.e. $f(gN)=f(g'N)$ whenever $gN=g'N$. In your case, \begin{align*} \Psi\colon G/({H\cap K}) & \longrightarrow G/H \times G/K \\ g(H\cap K) & \longmapsto (gH, gK), \\ \end{align*} take $g(H\cap K)=g'(H\cap K)$, so $g^{-1}g' \in H\cap K$. This means $g^{-1}g'\in H$ and $g^{-1}g' \in K$, so we have $gH=g'H$ and $gK=g'K$, so $(gH, gK)=(g'H, g'K)$ as desired.

Answer 2

Suppose $\phi : A\rightarrow B$, then to show $\phi$ is well defined you need to show that whenever $x,y\in A$ such that $x=y$, then $\phi(x)=\phi (y)$

Here $A=G/(H\bigcap K)$ and elements of this set are cosets of $(H\bigcap K)$ in $G$.

Elements of $A\times B$ are $(a,b)$ where $a\in A$ and $b\in B$. Now do you understand?

Note: The process is the "reverse" of what you'd do to show one-one.

It just means that the same element cannot map under $\phi$ to two different elements. Can you guess what would happen if the same elements mapped to two diferent elements under $\phi$? If you aren't able to figure out, read https://en.wikipedia.org/wiki/Function_(mathematics)

Answer 3

$$x(H\cap K)=y(H\cap K)\iff y^{-1}x\in H\cap K\subset\begin{cases}H\\K\end{cases}\implies$$

$$y^{-1}x\in H\;,\;\;K\implies\begin{cases} xH=yH\\xK=yK\end{cases}$$