Bounds on the index of an intersection of two subgroups

Let G be a group and let H,K \leq G be subgroups of finite index; say [G:H] = m and [G:K] = n. Prove that \mathsf{lcm}(m,n) \leq [G: H \cap K] \leq mn. Deduce that if m and n are relatively prime, then [G: H \cap K] = [G : H] \cdot [G : K].

Lemma 1: Let A and B be sets, \varphi : A \rightarrow B a map, and \Phi an equivalence relation on A. Suppose that if a_1 \,\Phi\, a_2 then \varphi(a_1) = \varphi(a_2) for all a_1,a_2 \in A. Then \psi : A/\Phi \rightarrow B given by [a]_\Phi \mapsto \varphi(a) is a function. Moreover, if \varphi is surjective, then \psi is surjective, and if \varphi(a_1) = \varphi(a_2) implies a_1 \,\Phi\, a_2 for all a_1,a_2 \in A, then \psi is injective. Proof: \psi is clearly well defined. If \varphi is surjective, then for every b \in B there exists a \in A such that \varphi(a) = b. Then \psi([a]_\Phi) = b, so that \psi is surjective. If \psi([a_1]) = \psi([a_2]), then \varphi(a_1) = \varphi(a_2), so that a_1 \,\Phi\, a_2, and we have [a_1] = [a_2]. \square

First we prove the second inequality.

Lemma 2: Let G be a group and let H,K \leq G be subgroups. Then there exists an injective map \psi : G/(H \cap K) \rightarrow G/H \times G/K. Proof: Define \varphi : G \rightarrow G/H \times G/K by \varphi(g) = (gH,gK). Now if g_2^{-1}g_1 \in H \cap K, then we have g_2^{-1}g_1 \in H, so that g_1H = g_2H, and g_2^{-1}g_1 \in K, so that g_1K = g_2K. Thus \varphi(g_1) = \varphi(g_2). Moreover, if (g_1H,g_2K) = (g_1H,g_2K) then we have g_2^{-1}g_1 \in H \cap K, so that g_1(H \cap K) = g_2(H \cap K). By Lemma 1, there exists an injective mapping \psi : G/(H \cap K) \rightarrow G/H \times G/K given by \psi(g(H \cap K)) = (gH,gK). \square

As a consequence, if [G : H] and [G : K] are finite, [G : H \cap K] \leq [G : H] \cdot [G : K].

Now to the first inequality.

Lemma 3: Let G be a group and K \leq H \leq G. Let S be a set of coset representatives of G/H. Then the mapping \psi : S \times H/K \rightarrow G/K given by \psi(g,hK) = ghK is bijective. Proof: (Well defined) Suppose h_2^{-1}h_1 \in K. Then h_1K = h_2K, so that gh_1K = gh_2K, and we have \psi(g,h_1K) = \psi(g,h_2K). (Surjective) Let gK \in G/K. Now g \in \overline{g}H for some \overline{g} \in S; say g = \overline{g}h. Then \psi(\overline{g},hK) = gK, so that \psi is surjective. (Injective) Suppose \psi(g_1,h_1K) = \psi(g_2,h_2K). Then g_1h_1K = g_2h_2K; in particular, g_1h_1 \in g_2h_2K \subseteq g_2H, so that g_1 \in g_2H and hence g_2^{-1}g_1 \in H. So g_1H = g_2H, and in fact g_2 = g_1. Thus h_1K = h_2K, and \psi is injective. \square

As a consequence, we have [G : H] \cdot [H : K] = [G : K].

Now in this case we have H \cap K \leq H \leq G. Thus m divides [G : H \cap K] and n divides [G : H \cap K], so that \mathsf{lcm}(m,n) divides [G : H \cap K]. In particular, since all numbers involved are natural, \mathsf{lcm}(m,n) \leq [G : H \cap K].

Finally, if m and n are relatively prime, then \mathsf{lcm}(m,n) = mn, and we have [G : H \cap K] = mn.

Post a comment or leave a trackback: Trackback URL.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: