I came across a proof earlier for a solution to a Herstein Topics in Algebra question earlier that I'm not convinced with, from AOPS site.
If $G$ is a group and $H,K$ are two subgroups of finite index in $G$, prove that $H \cap K$ is of finite index in $G$. Can you find an upper bound for the index of $H \cap K$ in $G$?
Since $[G:H]$ and $[G:K]$ are finite we can write:
$$G = Hx_{1} \cup Hx_{2} \cup \cdots \cup Hx_{r}$$
and
$$G = Ky_{1} \cup Ky_{2} \cup \cdots \cup Ky_{s}$$
where $[G:H]=r$ and $[G:K]=s$. Therefore,
\begin{align*}
G = G \cap G &= (Hx_{1} \cup Hx_{2} \cup \cdots \cup Hx_{r}) \cap (Ky_{1} \cup Ky_{2} \cup \cdots \cup Ky_{s}) \\
&= \bigcup_{\substack{1 \le i \le r \\ 1 \le j \le s}} (Hx_{i} \cap Ky_{j}).
\end{align*}
Hence, we need to consider the sets $Hx_{i} \cap Ky_{j}$ in the union. Suppose $Hx_{i} \cap Ky_{j} \neq \varnothing$ so that for some $g \in G$, $g \in Hx_{i} \cap Ky_{j}$. As right cosets in $G$ we have $Hg = Hx_{i}$ and $Hg = Ky_{j}$, so that we can write $Hg \cap Kg = Hx_{i} \cap Ky_{j}$. A small argument, set theoretical in nature, can show that $Hg \cap Kg = (H \cap K)g$. That is, $Hx_{i} \cap Ky_{j} = Hg \cap Kg = (H \cap K)g$ so each $Hx_{i} \cap Ky_{j}$ is either $\varnothing$ or some coset of $H \cap K$ in $G$. Since
$$G = \bigcup_{\substack{1 \le i \le r \\ 1 \le j \le s}} (Hx_{i} \cap Ky_{j})$$
is a finite union, $Hx_{i} \cap Ky_{j}$ is either $\varnothing$ or some coset of $H \cap K$ in $G$ we know every coset of $H \cap K$ appears in this union as cosets are either equal or disjoint and their union fills out all of $G$. We can conclude not only $[G:H \cap K]$ is finite but also
$$[G:H \cap K] \le [G:H][G:K]$$
since there are at most $[G:H][G:K]$ sets $Hx_{i} \cap Ky_{j}$.
The problem Im having is with the part saying
"Suppose $Hx_{i} \cap Ky_{j} \neq \varnothing$ so that for some $g \in G$, $g \in Hx_{i} \cap Ky_{j}$. As right cosets in $G$ we have $Hg = Hx_{i}$ and $Hg = Ky_{j}$, so that we can write $Hg \cap Kg = Hx_{i} \cap Ky_{j}$"
I understand that any two cosets of the same subgroup are either equal or disjoint, but why should this hold for 2 different subgroups?
e.g. consider the integers under addition. ${4n}, {3n}$ are subgroups but the cosets ${4n+1}, {3n+1}$ only have some elements in common.
So if someone could explain why/if we can conclude $Hx_{i}$, $Ky_{j}$ must equal the same coset $Hg$ it would be appreciated.
(I understand alternative solutions exist to this on MSE but I was wandering about this proof specifically).
Best Answer
Let $g \in Hx_i \cap Ky_j$. Then there are $h \in H$ and $k \in K$ such that $g= hx_i = ky_j$. Hence we have $$Hg = H(hx_i) = (Hh)x_i = Hx_i$$ and similarly $Kg = K(ky_j) = Ky_j$. And therefore $$Hg \cap Kg = Hx_i \cap Kx_j.$$