Exercise 3.2.10 (If $\mathrm{g.c.d}(|G:H|, |G:K|) = 1$, then $|G : H \cap K| = |G : H | \cdot |G: K|$)

Question

Suppose that $H$ an $K$ are subgroups of finite index in the (possibly infinite) group $G$ with $|G:H| = m$ and $|G:K| =n$. Prove that $\mathrm{l.c.m.}(m,n)
 \leq |G : H \cap K| \leq mn$. Deduce that if $m$ and $n$ are relatively prime then $|G : H \cap K| = |G : H | \cdot |G: K|$.

Richard Ganaye · Accepted Answer

Note: We must solve Exercise 11 first.

\begin{proof} We first prove (in the spirit of the proof of Proposition 13)  that
\begin{align}
|H : H \cap K | \leq |G : K|.
\end{align}
For this sake, consider the set of left cosets $h K$, where $h \in H$.
$$S = \{h K \mid h \in H\}.$$
Then $S \subseteq \{g K \mid g \in G\}$, where the set $\{g K \mid g \in G\}$ of all left cosets of $K$ in $G$ has $|G : K|$ elements, thus 
$$|S| \leq |G : K|,$$
so $S$ is a finite set.

Consider the map
$$\varphi \ 
\left\{
\begin{array}{ccl}
H & 	o & S\
h & \mapsto & hK.
\end{array}≤
ight.
$$
For all $h, h' \in H$,
\begin{align*}
\varphi(h) = \varphi(h') & \iff h K = h' K\
&\iff h'^{-1} h  \in H \cap K\
&\iff h (H \cap K) = h' (H \cap K).
\end{align*}
Therefore the preimage by $\varphi$ of any $hK \in S$ has $|H : H \cap K|$ elements, and ${|S| = |H : H \cap K|}$. Since $|S| < \infty$, this shows that  $|H : H \cap K| < \infty$, and the inequality (1) is proved.

\bigskip
By Exercise 11, since $|G : H|$ and $|H : H \cap K|$ are finite,
\begin{align}
|G : H \cap K| = |G : H| \cdot |H : H \cap K|,
\end{align}
where $|H : H \cap K | \leq |G : K|$, therefore
$$|G : H \cap K| \leq |G : H| \cdot |G : K| = mn.$$

\bigskip
Moreover the equality (2) shows that $m = |G : H|$ divides $|G : H \cap K|$, and similarly $|G : H \cap K| = |G : K| \cdot |K : H \cap K|$ shows that $n$ divides $|G : H \cap K|$.

Let $m \vee n$ denote $\mathrm{l.c.m.}(m,n)$ and $m \wedge n = \mathrm{g.c.d}(m,n)$. Then, if $ i =  |G : H \cap K|$,
$$(m \mid  i	ext{ and } n \mid i )\Rightarrow m \vee n \mid i.$$
Therefore $n \vee m \leq |G : H \cap K|$, so
$$n \vee m \leq |G : H \cap K| \leq mn.$$

If $n \wedge m = 1$, then $n \vee m = nm$ (recall that $(n \wedge m) (n \vee m) = nm$). Therefore $nm \leq |G : H \cap K| \leq nm$, so
$$|H|  \wedge |K| = 1 \Rightarrow |G : H \cap K|  = |G:H|\cdot |G:K|.$$
\end{proof}

Exercise 3.2.10 (If $\mathrm{g.c.d}(|G:H|, |G:K|) = 1$, then $|G : H \cap K| = |G : H | \cdot |G: K|$)

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Exercise 3.2.10 (If $\mathrm{g.c.d}(|G:H|, |G:K|) = 1$, then $|G : H \cap K| = |G : H | \cdot |G: K|$)

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