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Let $G$ be a group and $H,K$ subgroups of $G$. Furthermore, let $O_H=\{aH|a\in G\},O_K=\{aK|a\in G\}$ and $O_{H\cap K}=\{aH\cap K|a\in G\}$.

I was able to prove that $f:O_{H\cap K}\rightarrow O_H\times O_K$, $f(aH\cap K)=(aH,aK)$ is well-defined and onto.

Now I want to prove: if $[G:H]<\infty$ and $[G:K]<\infty$, then $[G:(H\cap K)]<\infty$. But how to do it? I only know that by Lagrange Theorem it holds $[G:H]=|G|/|H|$, $[G:K]=|G|/|K|$ and also $[G:(H\cap K)]=|G|/|H\cap K|$.

  • By the way, there's a standard notation for the set of cosets of a subgroup $H$ in group $G$, regardless of whether the subgroup is normal so the cosets form a group themselves: $G/H$. – Sammy Black Nov 25 '24 at 18:48
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    @SammyBlack: Some notational care is needed in that the "set of cosets" is not defined when the subgroup is not normal. There is a "set of left cosets" denoted $$G , / , H = {aH \mid a \in G}$$ and there is a "set of right cosets" denoted $$H\setminus G = {Hb \mid b \in G}$$ – Lee Mosher Nov 25 '24 at 18:52
  • Yes, of course. – Sammy Black Nov 25 '24 at 18:53

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