3
$\begingroup$

$H, K$ 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!

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – Erick Wong
    Jan 5 '13 at 3:26
  • $\begingroup$ @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? $\endgroup$ Jan 5 '13 at 3:32
  • 1
    $\begingroup$ 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 $H \cap K$). $\endgroup$
    – Erick Wong
    Jan 5 '13 at 3:35
  • $\begingroup$ @Erick Wong: Ah, correct. Mind copying and pasting your comments as an answer, so I can accept (even if it strikes you as trivial)? $\endgroup$ Jan 5 '13 at 3:41
  • $\begingroup$ No objections here :). $\endgroup$
    – Erick Wong
    Jan 5 '13 at 3:42
6
$\begingroup$

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 $zH \cap zK \subseteq z(H \cap K).$

$\endgroup$
1
  • $\begingroup$ Very nice. Thank you. $\endgroup$ Jan 5 '13 at 16:34
2
$\begingroup$

As requested, I'm turning the comments into an answer:

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 $H\cap K$).

$\endgroup$

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.