Let \(G\) be a finite group and \(H<G\). If \([G:H]=2\), then \(gH=Hg\) for any \(\in G\).
Proof: If \([G:H]=2\), then there are only two cosets of \(H\) in \(G\), and one of the cosets is \(H\) itself, i.e.,

for some \(g_0\in G\), where \(H\cap g_0H=\varnothing,~H\cap Hg_0=\varnothing\). For any \(g\in G\). If \(g\in H\), it is obviously that \(gH=H=Hg\). Else if \(g\notin H\), then \(gH\neq H,~Hg\neq H\). Since \([G:H]=2\), we have \(G=H\cup gH=H\cup Hg\) and \(H\cap gH=\varnothing,~H\cap Hg=\varnothing\). Thus \(gH=Hg\).

Suppose that \([G:H]=2\). If \(a,b\notin H\), show that \(ab\in H\).
Proof: If \([G:H]=2\), then \(G=H\cup gH\) and \(H\cap gH=\varnothing\). If \(a,b\notin H\), then \(a,b\in gH\), i.e., \(\exists h_1,h_2\in H\), such that \(a=gh_1,~b=gh_2\). Then \(ab=gh_1gh_2\in G\). Assume that \(ab\notin H\), then \(ab\in gH\), i.e., \(\exists h_3\in H\), s.t. \(ab=gh_3\), i,e, \(gh_1gh_2=gh_3\), then \(h_1gh_2=h_3\Rightarrow g=h_1^{-1}h_3h_2^{-1}\in H\Rightarrow gH=H\). Since \(H\cap gH=\varnothing\), we have \(H=gH=\varnothing=G\), it contradicts to the fact that \([G:H]=2\). Thus, \(ab\in H\).

Let \(H,K<G\) and \(|H|=12,|K|=35\). What is \(H\cap K\).
Solution: Since \(H\cap K<H\) and \(H\cap K<K\), we have \(|H\cap K|\mid |H|\) and \(|H\cap K|\mid |K|\) by the Lagrange's Theorem. Note that \(\gcd(12,35)=1\), so \(|H\cap K|=1\Rightarrow H\cap K=\{\text{Identity}\}\).