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## Bounds on the index of an intersection of two subgroups

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.10

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 $[G: H \cap K] = [G : H] \cdot [G : K]$.

Solution:

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 $[G : H \cap K] = mn$.