## A problem from Artin's algebra textbook

1. The problem statement, all variables and given/known data
(a)Let H and K be subgroups of a group G. Prove that the intersection of xH and yK which are cosets of H and K is either empty or else is a coset of the subgroup H intersect K

(b) Prove that if H and K have finite index in G then the intersection of H and K also has finite index.

2. Relevant equations

3. The attempt at a solution
The intersection of xH and yK is a subgroup of both H and K, then how to continue?

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 Quote by AbelAkil 3. The attempt at a solution The intersection of xH and yK is a subgroup of both H and K, then how to continue?
This is not true in general. If xH and yK are not subgroups, then neither contains the identity, so their intersection also doesn't contain the identiy. So it can't be a subgroup.

Moreover, in general $xH \cap yK$ isn't even a subSET of H or K. xH and H are disjoint unless $x \in H$. Similarly for yK and K.

 Quote by jbunniii This is not true in general. If xH and yK are not subgroups, then neither contains the identity, so their intersection also doesn't contain the identiy. So it can't be a subgroup. Moreover, in general $xH \cap yK$ isn't even a subSET of H or K. xH and H are disjoint unless $x \in H$. Similarly for yK and K.
Sorry, I made some mistakes when I wrote the post. In fact, I mean the intersection of H and K is a subgroup of both H and K...Could U give me some tips to prove it?

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## A problem from Artin's algebra textbook

If xH and yK have nonempty intersection, then there is an element g contained in both: $g \in xH$ and $g \in yK$.

The cosets of $H \cap K$ form a partition of G, so g is contained in exactly one such coset, call it $a(H \cap K)$.

If you can show that $a(H \cap K)$ is contained in both $xH$ and $yK$ then you're done.

Hint: both $xH$ and $yK$ are partitioned by cosets of $H \cap K$.

 Quote by jbunniii If xH and yK have nonempty intersection, then there is an element g contained in both: $g \in xH$ and $g \in yK$. The cosets of $H \cap K$ form a partition of G, so g is contained in exactly one such coset, call it $a(H \cap K)$. If you can show that $a(H \cap K)$ is contained in both $xH$ and $yK$ then you're done. Hint: both $xH$ and $yK$ are partitioned by cosets of $H \cap K$.
Yeah...I get it. Thanks very much. In addition, how to prove part (b), that is how can I show that both $H$ and $K$ are partitioned by finite cosets of $H \cap K$... I appreciate your insightful answer!
 Recognitions: Science Advisor the index of H in G is the number of cosets of H. if this number is finite, then if it just so happened that H∩K was of finite index in H, we get: [G:H][H:H∩K] cosets of H∩K in G in all, which would be finite. can you think of a way to show that [H:H∩K] ≤ [G:K]? perhaps you can think of an injection from left cosets of H∩K in H to left cosets of K in G?

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