MHB How can we prove the equality of orders for finite subgroups of a group?

[mathmari]

Hey!

I want to show that if $H, K\leq G$ are finite subgroups of $G$, then $|HgK|=\frac{|H||K|}{|K\cap H^g|}$ where $g\in G$.

We have that $H^g=g^{-1}Hg$ and $|HK|=\frac{|H||K|}{|H\cap K|}$.

So we have that $|HgK|=|gH^gK|=\frac{|gH^g||K|}{|(gH^g)\cap K|}$, or not? (Wondering)

What how can we get the desired result? (Wondering)

 

[Deveno]

It's a counting formula.

So basically, you want to count the elements of the set:

$HgK = \{hgk: h\in H, k \in K\}$.

We can think of this as the PRODUCT set:

$(Hg)K = \{ak: a\in Hg, k \in K\}$.

We know that the product set is bounded by $|Hg|\cdot|K| = |H|\cdot|K|$.

So the question becomes: which elements do we count more than once?

Suppose $h_1gk_1 = h_2gk_2$.

This means $h_1g = h_2gk_2k_1^{-1}$.

Since $K$ is a subgroup, $k_2k_1^{-1} \in K$, let's call it $k'$.

So we have: $h_1g = h_2gk'$. Continuing:

$g^{-1}h_2^{-1}h_1g = k'$.

Since $H$ is a subgroup, $h_2^{-1}h_1 \in H$, say, it's $h'$.

Then $g^{-1}h'g = k'$, that is, if $x = k'$, we have $x \in H^g$ and $x \in K$, so $x \in (H^g) \cap K$.

I sincerely hopes this convinces you that if we *do* count something more than once, it is *because* we have something in the intersection of $H^g$ and $K$.

So let's turn this around, now, suppose we have $x = g^{-1}hg=k$.

Then $g = h^{-1}gk$. If the elements of $H^g \cap K$ are:

$g^{-1}h_1g = k_1$
$g^{-1}h_2g = k_2$
$\vdots$
$g^{-1}h_ng = k_n$

This gives us $n = |H^g \cap K|$ ways to write the element $g \in HgK$, as $g = h_j^{-1}k_j$, for $j = 1,\dots,n$.

Similarly, for any other element:

$hgk \in HgK$, we can re-write this as:

$hgk = h(h_j^{-1}gk_j)k$ in $n$ different ways (one for each $j$).

This means that each element of $HgK$ is counted exactly $n$ times:

$|HgK| = \dfrac{|Hg|\cdot|K|}{n} = \dfrac{|Hg|\cdot|K|}{|H^g\cap K|} = \dfrac{H|\cdot|K|}{|H^g\cap K|}$.

For an extra bonus, what happens if the element $g=e$?

 

[mathmari]

We know that the product set is bounded by $|Hg|\cdot|K| = |H|\cdot|K|$.

We have that $Hg=\{hg\mid h\in H\}$, so the number of elements of $Hg$ is equal to the number of $H$, since it is a right product of each element of $H$ with a fixed $g$, right? (Wondering)

This means that each element of $HgK$ is counted exactly $n$ times:

$|HgK| = \dfrac{|Hg|\cdot|K|}{n} = \dfrac{|Hg|\cdot|K|}{|H^g\cap K|} = \dfrac{H|\cdot|K|}{|H^g\cap K|}$.

When we have that each element is counted exactly $n$ times, we do we divide $ |Hg|\cdot|K|$ by $n$ and not substract it by $n$ ? (Wondering)

 

[Deveno]

With natural numbers, at least, "times" means multiplication, not addition.

 

[mathmari]

I got stuck right now... I haven't really understood why we divide by $n$. Could you explain it further to me? (Wondering)

 

[Deveno]

I got stuck right now... I haven't really understood why we divide by $n$. Could you explain it further to me? (Wondering)

For each $x \in H^g \cap K$, we have $n = |H^g \cap K|$ ways to write $hgk$, for any pair $(h,k)$ where $h \in H, k\in K$.

But here is a slicker proof I thought of the other day:

IF you already know that:

$|HK| = \dfrac{|H|\cdot|K|}{|H \cap K}|$ for any two subgroups, $H,K$ of $G$ ($\ast$),

then using the fact that for any SET $S \subseteq G$, the map $S \to g^{-1}S = \{g^{-1}s| s\in S\}$ given by:

$s \mapsto g^{-1}s$ is a bijective map,

we have $|HgK| = |g^{-1}(HgK)| = |H^gK|$

and since $H^g$ is a subgroup of $G$ with $|H^g| = |H|$, we have:

$|HgK| = |H^gK| = \dfrac{|H^g|\cdot|K|}{|H^g\cap K|} = \dfrac{|H|\cdot|K|}{|H^g\cap K|}$

by cleverly substituting $H^g$ for $H$ in our formula ($\ast$).

 

[mathmari]

and since $H^g$ is a subgroup of $G$ with $|H^g| = |H|$

We have that $|H^g| = |H|$, because the map $H\rightarrow H^g=g^{-1}Hg=\{g^{-1}hg\mid h\in H\}$ given by $h\mapsto g^{-1}hg$ is bijective, right? (Wondering)

 

[Deveno]

We have that $|H^g| = |H|$, because the map $H\rightarrow H^g=g^{-1}Hg=\{g^{-1}hg\mid h\in H\}$ given by $h\mapsto g^{-1}hg$ is bijective, right? (Wondering)

Indeed. In fact, one can view it as the *composition* of the two bijective maps:

$g \mapsto hg$ and $x \mapsto g^{-1}x$, with $x = hg$.

Why are these maps bijective? Essentially, we can "undo" each one by multiplying (on the appropriate side) by an appropriate inverse. That is, in a nutshell, what groups actually ARE: sets of composable and reversible mappings (the set the mappings act on may well be the underlying set of the group itself, but it may be something else, like a finite set of objects, or a geometric figure).

 

[mathmari]

Ah ok...

Thank you very much! (Mmm)