# $xH \cap yK$ is either empty or a coset of $H \cap K$: blanking

$H, K$ are sub groups of a group $G$. What I have:

Denote the left cosets $H_i$, $K_j$, and $(H \cap K)_k$, respectively. If $x, y \in H_i \cap K_j$, then $y^{-1}x \in H, K$, and so in $H \cap K$, so $x, y$ are in the same coset $(H \cap K)_k$ for some $k$.

Where I'm blanking: what if $H_i \cap K_j$ has only one element? How to proceed then, and so conclude the proof?

Thanks!

-
Your argument applies no matter how many elements $H_i \cap K_j$ has, so I see no problem there. However, you've only shown that $H_i \cap K_j$ is contained in some coset of $(H \cap K)$, and you haven't said anything about why it covers the entire coset. – Erick Wong Jan 5 '13 at 3:26
@Erick Wong: Good point on the second, and that shouldn't be so hard. But I don't see what you say first: why does it not matter? I have, say, $a = xh = yk$...and then? You are probably right, but could you elaborate? – gnometorule Jan 5 '13 at 3:32
This half of the argument is just trying to show that all of $H_i \cap K_j$ is contained in one coset of $H \cap K$. You've already showed that any two elements belong to the same coset. If there's only one element, there's nothing more to prove (of course it belongs to its own coset of $H \cap K$). – Erick Wong Jan 5 '13 at 3:35
@Erick Wong: Ah, correct. Mind copying and pasting your comments as an answer, so I can accept (even if it strikes you as trivial)? – gnometorule Jan 5 '13 at 3:41
No objections here :). – Erick Wong Jan 5 '13 at 3:42

Your argument applies no matter how many elements $H_i \cap K_j$ has, so I see no problem there. However, you've only shown that $H_i \cap K_j$ is contained in some coset of $H\cap K$, and you haven't said anything about why it covers the entire coset.
In any case, this half of the argument is just trying to show that all of $H_i \cap K_j$ is contained in one coset of $H\cap K$. You've already shown that any two elements belong to the same coset. If there's only one element, there's nothing more to prove (of course it belongs to its own coset of $H\cap K$).
The question is already answered, but note that if $xH \cap yK$ is non-empty, then it contains an element $z$. We have $xH = zH$ and $yK = zK,$ so it suffices to understand the case $x = y = z.$ Now $z(H \cap K) \subseteq zH \cap zK,$ so $zH \cap zK$ contains at least one full coset of $H \cap K.$ On the other hand, if $zh = zk$ for some $h \in H$ and $k \in K,$ then $h = k \in H \cap K$, so $zH \cap zK \subseteq z(H \cap K).$