The Intersection of Two Subgroups of Finite Index Has Finite Index

Question

I am attempting to work through Shatz and Gallier's Algebra, and I would like help verifying and cleaning up the proof of the following problem:

Problem: In an infinite group, prove that the intersection of two subgroups of finite index has finite index itself.

Proof: Let $H, K$ be subgroups of a group $G$. It suffices to show that $(G:H \cap K)\leq (G:H)(G:K).$ Let $C = \{g_iH\cap K\}, C' = \{g_j'H\}, C'' = \{g_k''K\}$ be the sets of distinct left cosets of $H\cap K, H$ and $K$ respectively. Consider the map \begin{align*} \varphi: C&\hookrightarrow C'\times C'' \\ g_iH\cap K &\mapsto (g'_{j_i}H,g''_{k_i}K) \end{align*} where $g'_{j_i}$ and $g''_{k_i}$ are such that $g_iH\cap K \subset g'_{j_i}H$ and $g_iH\cap K \subset g''_{k_i}K.$ Check that this map is well-defined: there cannot be two distinct cosets $g'_{j_i}H, g'_{j'_i}H$ containing $g_iH\cap K$ as a subset, so we need only make sure $g'_{j_i}H$ exists. Choose an element $g_il$ of $g_iH\cap K$ and label the coset in $C'$ in which it lies $g'_{j_i}H.$ Then $g_il = g'_{j_i}h$ for some $h\in H,$ and for any $l' \in H\cap K$ $g_il' = g'_{j_i}hl^{-1}l'\in g'_{j_i}H.$ Thus the entire coset $g_iH\cap K$ lies in $g'_{j_i}H.$ By an analogous argument there exists a unique $g_k''K$ such that $g_iH\cap K \subset g''_{k_i}K.$

We now show that $\varphi$ is an injection. Suppose $g_1H\cap K, g_2H\cap K$ both lie in the same cosets $g'H, g''K.$ Fix $l \in H\cap K.$ Then $g_1l = g'h_1 = g''k_1$ for some $h_1, k_1 \in H, K$ respectively. Similarly $g_2l = g'h_2 = g''k_2$ for $h_2, k_2 \in H, K$ respectively. Then \begin{align*} g_1l &= g'h_1 = g''k_2h_2^{-1}h_1 = g_2lh_2^{-1}h_1\in g_2H \text{ and }\\ g_1l &= g''k_1 = g'h_2k_2^{-1}k_1 = g_2lk_2^{-1}k_1\in g_2K \end{align*} and after combining these statements we have that $g_1l \in g_2H\cap K,$ and $g_1, g_2$ define the same coset w.r.t. $H\cap K.$

Is this proof correct?

Answer 1

I like to offer an alternative proof, which is less computational. Let $H,K$ be subgroups of $G$ with finite indices. The action of $G$ on the set of left cosets $G/H$ by left multiplication yields a homomorphismus $f:G\to\mathrm{Sym}(G/H)$ with kernel $H_G\le H$ (the core of $H$ in $G$). Note that $H_G\unlhd G$ and $|G:H_G|\le|G:H|!$ is finite. It is easy to see that $G\to G/H_G\times G/K_G$, $g\mapsto (gH_G,gK_G)$ is homomorphism with kernel $H_G\cap K_G$. In particular, $|G:H_G\cap K_G|\le|G:H_K||G:K_G|$ is finite. The claim follows since $H_G\cap K_G\le H\cap K$.

Answer 2

I think your argument can be written far more succinctly.

I use (some of) your notation.

We want to define a monomorphism $\phi:C\to C'\times C''$.

Define $\phi( x(H\cap K))=(xH,xK)$. This is well-defined since if we have $x(H\cap K)=y(H\cap K)$ then $x(H\cap K)\subseteq xH$ and $y(H\cap K)\subseteq yH$ and so $xH\cap yH$ is non-empty, so $xH=yH$. The same argument applies to $K$.

It is also true that $\phi$ is one-to-one. For if $(xH,xK)=(yH,yK)$ then $xH=yH$ so that $xy^{-1}\in H$. Similarly $xy^{-1}\in K$. Hence $xy^{-1}\in H\cap K$ and thus $x(H\cap K)=y (H\cap K)$.