Let \(G\) be a group and let \(H,K\le G\). Suppose \(H\) has finite index in \(G\). Prove that \[ K\cap H \] has finite index in \(K\).
Solution
1. Define\[ \Phi:K/(K\cap H)\longrightarrow G/H \]on left cosets by\[ \Phi\bigl(k(K\cap H)\bigr)=kH. \]
This gives a candidate map from the set of left cosets of \(K\cap H\) in \(K\) to the set of left cosets of \(H\) in \(G\).
2. The map \(\Phi\) is well-defined.
Suppose \[ k_1(K\cap H)=k_2(K\cap H). \] Then \[ k_2^{-1}k_1\in K\cap H\subseteq H, \] so \[ k_1H=k_2H. \] Thus the value of \(\Phi\) depends only on the coset.
3. The map \(\Phi\) is injective.
Suppose \[ \Phi\bigl(k_1(K\cap H)\bigr) = \Phi\bigl(k_2(K\cap H)\bigr). \] Then \(k_1H=k_2H\), so \[ k_2^{-1}k_1\in H. \] Since \(k_1,k_2\in K\), one also has \(k_2^{-1}k_1\in K\). Hence \[ k_2^{-1}k_1\in K\cap H, \] which means \[ k_1(K\cap H)=k_2(K\cap H). \]
4. The index \([K:K\cap H]\) is finite.
By step 3, \[ [K:K\cap H] = \abs{K/(K\cap H)} \le \abs{G/H} = [G:H]. \] The right-hand side is finite by hypothesis. Therefore \[ \boxed{[K:K\cap H]<\infty}. \]
5. Q.E.D.
Step 4 is the required conclusion.