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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).

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    $\begingroup$ I'm wondering if there's a typo: Should it rather say "As right cosets in G we have $Hg = Hx_i$ and $Kg = Ky_j$" (i.e. change $H$ to $K$ in the second equality)? $\endgroup$ Jul 23 '16 at 11:26
  • $\begingroup$ @TomBombadil Ah yea thats probably a typo, but then why must the '$g$s' be the same, i.e. why cant $Hg_{1} = Hx_{i}$ and $Kg_{2} = Ky_{j}$ for $g_{1} \not= g_{2} $ ? $\endgroup$ Jul 23 '16 at 11:32
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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 $$Hg \cap Kg = Hx_i \cap Kx_j.$$

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  • $\begingroup$ Ah yes of course. So would you say this proof is valid after that correction then? $\endgroup$ Jul 23 '16 at 11:36
  • $\begingroup$ Yes, after correcting this small typo, the above proof is valid. $\endgroup$ Jul 23 '16 at 11:41

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