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Let $H$, $K$ be subgroups of a given group $G$. Can one show that $(G:(H\cap K))$ is less or equal to $(G:H)(G:K)$, where $(G:H)$ stands for the index of group G with respect to $H$?

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$(G:(H\cap K)) = (G:H)(H:H\cap K)$. So this amounts to showing that $(H:H\cap K)\leq (G:K)$ – Thomas Andrews Oct 28 '11 at 20:33
(Note the simplest route, just pointing out a related post.) This question asks to prove that $|H| |K| \leq |H \cap K| | \langle H, K \rangle |$, which is equivalent to $(G:H)(G:K) \geq (G:H \cap K) (G: \langle H, K \rangle)$. Since $(G: \langle H, K \rangle) \geq 1$, we get the claim in this question (assuming that question). – Srivatsan Oct 28 '11 at 20:33
@Thomas, indeed but this is my main problem. Thank you. – Ticuramamba Oct 28 '11 at 20:36
I probably make a horrible mistake, but consider the function $ \pi : H \hookarrow G \rightarrow G/K$, as a function on cosets. This should induce an injection $f: H/(H\cap K) \rightarrow G/K$ which proves your statement. If all subgroups are normal, I think this idea works imediately, my sugestion is to forget the group structure of $H/(H\cap K)$ and $G/K$ and work set theoretically. Just be careful if you use left or right cosets. – N. S. Oct 28 '11 at 21:42
up vote 1 down vote accepted

Proposition. If $H$ and $K$ are subgroups of $G$, then $[H:H\cap K]\leq [\langle H,K\rangle:K]$ in the sense of cardinalities. If $\langle H,K\rangle = HK$, then we have equality; in the finite case, this condition is also necessary for equality.

Proof. Let $\mathcal{C}$ be the set of all left cosets of $H\cap K$ in $H$, and let $\mathcal{D}$ be the set of all left cosets of $K$ in $\langle H,K\rangle$.

Define a (set-theoretic) map $f\colon\mathcal{C}\to\mathcal{D}$ by $f(h(H\cap K)) = hK$.

First, this is well-defined and injective: if $h,h'\in H$, then $$\begin{align*} h(H\cap K) = h'(H\cap K) &\Longleftrightarrow h^{-1}h'\in H\cap K\\ &\Longleftrightarrow h^{-1}h'\in K\\ &\Longleftrightarrow hK = h'K. \end{align*}$$ Moreover, note that since $H\subseteq \langle H,K\rangle$, then $hK\in\mathcal{D}$.

Therefore, $f$ is a one-to-one function from $\mathcal{C}$ to $\mathcal{D}$, hence $[H:H\cap K] = |\mathcal{C}| \leq |\mathcal{D} = [\langle H,K\rangle:K]$.

If $\langle H,K\rangle = HK$, then every element of $\mathcal{D}$ is of the form $hK$ for some $h\in H$, so $f$ will also be surjective, giving $|\mathcal{C}|=|\mathcal{D}|$. In the finite case, if we have equality, then since $f$ is one-to-one between two finite sets of the same size, it must be onto, so given any element $g$ of $\langle H,K\rangle$, there exists $h\in H$ such that $gK=hK$; hence there exists $k\in K$ such that $g=hk\in HK$, so $\langle H,K\rangle\subseteq HK$, as desired. $\Box$

Applying this to your case, we have that in the sense of cardinalities, $$\begin{align*} [G:H\cap K] &= [G:H][H:H\cap K] \\ &\leq [G:H][\langle H,K\rangle:K]\\ &\leq [G:H][G:K] \end{align*}$$ If $G$ is finite, then equality holds if and only if $HK=G$.

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You could show that $a(H \cap K) = aH \cap aK\ $ for every $a\in G$. We can choose $aH \cap aK$ in $[G:H][G:K]$ ways. Some of the combinations might be same, but there can be no more than $[G:H][G:K]$ cosets of $H \cap K$.

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I feel like it's worth mentioning that if $G$ is finite (probably true more generally) and $H\leqslant N_G(K)$ then $[G:H\cap K]\mid [G:H][G:K]$. Indeed, by containment in the normalizer we know that $HK\leqslant G$ and so $\displaystyle \frac{|G|}{|HK|}$ is an integer. That said,

$\displaystyle \begin{aligned}\frac{|G|}{|HK|}&= \frac{|G|}{\displaystyle \frac{|H||K|}{|H\cap K|}}\\ &= \frac{\displaystyle \frac{|G|}{|H|}\frac{|G|}{|K|}}{\displaystyle \frac{|G|}{|H\cap K|}}\\ &= \frac{[G:H][G:K]}{[G:H\cap K]}\end{aligned}$

I don't know if that's relevant to your interests, but I thought it's worth mentioning. It helps give a proof why the only subgroup of $S_n$ (for $n\geqslant 5$) of index $2$ is $A_n$.

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Well, for divisibility to make sense to you need the indices to be finite; so you need $H$ and $K$ to be finite index in $G$ before the expression makes sense. If $H$ and $K$ have finite index, then so does $H\cap K$, and so does the core of $H\cap K$ (the largest normal subgroup that is contained in $H\cap K$). Moding out by the core of $H\cap K$, you can reduce to the finite case. – Arturo Magidin Oct 29 '11 at 22:09

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