# The Intersection of Two Subgroups of Finite Index Has Finite Index

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?

• Don't you mean $x(H\cap K)$ every time you write $xH\cap K$? Commented Sep 16, 2022 at 7:23
• I believe that it is correct. (I also believe it's a notational nightmare and can be written far more compactly, and without all these sub- and superscripts.) Commented Sep 16, 2022 at 7:29
• I agree. I am asking for help making it clearer. How would I do that? Commented Sep 16, 2022 at 12:32

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 $$H_G\cap K_G\le H\cap K$$.

• I agree this is the best way to think of it. Commented Sep 16, 2022 at 13:49

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 $$x(H\cap K)=y (H\cap K)$$.