Why does injectivity imply to $|G/(H \cap K)|\leq |G/H|\cdot|G/K|?$

Question

I'm looking through some proof about the inequality in the title, the one defines:

$$\phi: G/(H \cap K)\rightarrow G/H\times G/K$$

$$\phi(g(H \cap K))=(gH,gK)$$

Note that $\phi$ is injective, I'd like to know why the such injectivity implies that:

$$\left|G/(H \cap K)\right| \leq |G/H|\cdot|G/K|$$

I have no idea, you don't need to prove just getting a useful idea to answering this is enough.

If $S$ and $T$ are finite sets and $f: S \to T$ is an injective function, do you see why $|S| \leq |T|$? — Daniel, Apr 12 '22 at 21:31
@Daniel, I guess no, because this is precisely what the OP is asking. — , Apr 12 '22 at 21:44
Does this answer your question? Proof by induction regarding injections and part of the pigeonhole principle — , Apr 12 '22 at 22:01

score 2 · Accepted Answer · answered Apr 12 '22 at 21:33

Note that $\left \vert \frac GH \times \frac GK \right \vert = \left \vert \frac GH \right \vert \cdot \left \vert \frac GK \right \vert$. Since $\phi$ is injective, every element of $\frac {G}{H \cap K}$ maps to a unique ordered pair in the cross product $\frac GH \times \frac GK$ (possibly with some ordered pairs in the cross product left over). This proves the desired inequality because you can't injectively map a set into a smaller set.

Why does injectivity imply to $|G/(H \cap K)|\leq |G/H|\cdot|G/K|?$

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