The question is from Grillet's lovely Abstract Algebra:

Let $A,B$ be two cosets of possibly different normal subgroups of finite index of a group $G$. It is to be shown that $A \cap B$ is either empty or the coset of a normal subgroup of $G$, also of finite index.

So far I have assumed that the intersection is not empty and that the normal subgroups are different. In this case, we can write:

$xN \cap yH = \{ z: \exists n \in N, h \in H : z=xn=yh \}$

where $N,H$ are our normal subgroups of finite index and $x,y \in G$ fixed. I haven't really gotten further. I've also tried applying the second isomorphism theorem but it doesn't seem to stick. A hint or some guidance would be welcome.


$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$Suppose $N, H$ are any subgroups of finite index in $G$.

If $x N \cap y H \ne \emptyset$, let $z \in x N \cap y H$. Then $z \in x N$, so that $z N = x N$, and similarly $z H = y H$, so $$ x N \cap y H = z N \cap z H = z (N \cap H). $$ In fact clearly $z (N \cap H) \subseteq z N \cap z H$, and if $w \in z N \cap z H$, then $w = z n = z h$ for some $n \in N$ and $h \in H$, and thus $n = h \in N \cap H$.

Now use the fact that if two subgroups have finite index, then their intersection has finite index. This follows from the formula $$ \Size{N H : H} = \Size{N : N \cap H} $$ which holds if $H$ has finite index (note that $N H$ is not necessarily a subgroup), so that if $N$ has also finite index $$ \Size{G : N \cap H} = \Size{G : N } \cdot \Size{N : N \cap H}. $$

  • $\begingroup$ This is such a beautiful answer, thank you. $\endgroup$ – ZirconCode Jan 2 '17 at 13:38
  • 1
    $\begingroup$ @ZirconCode, you're very welcome! $\endgroup$ – Andreas Caranti Jan 2 '17 at 13:39

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.