Order of Subgroup Product/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $G$ be a group.

Let $H$ and $K$ be subgroups of $G$.


Then:

$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$

where:

$H K$ denotes subset product
$\order H$ denotes the order of $H$.


The validity of the material on this page is questionable.
In particular: The right hand side does not make sense if $\order {H \cap K} = +\infty$, because it is then $\dfrac {+\infty}{+\infty}$. Can someone check the source?
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Questionable}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



Proof

Lemma

Let $h_1, h_2 \in H$.

Then:

$h_1 K = h_2 K$

if and only if:

$h_1$ and $h_2$ are in the same left coset of $H \cap K$.

$\Box$

We have that $H K$ is the union of all left cosets $h K$ with $h \in H$:

$\ds H K = \bigcup_{h \mathop \in H} h K$

From Left Coset Space forms Partition, unequal $h K$ are disjoint.

From Cosets are Equivalent, each $h K$ contains $\order K$ elements.


From the Lemma, the number of different such left cosets is:

$\index H {H \cap K}$

where $\index H {H \cap K}$ denotes the index of $H \cap K$ in $H$.


First, let $\order H < + \infty$.

Then, from Lagrange's Theorem:

$\index H {H \cap K} = \dfrac {\order H} {\order {H \cap K} }$

Hence:

$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$


Finally, let $\order H = + \infty$.

Recall that $G$, $H$ and $K$ have the same identity element $e$ by Identity of Subgroup.

By Definition of Subset Product:

$H = H \set e \subseteq H K$

In particular, $\order {H K} = + \infty$,

Hence:

$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} } = + \infty$


This needs considerable tedious hard slog to complete it.
In particular: How to handle the case $\order {H \cap K} = +\infty$? The statement of the theorem leaves the ambiguity.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Order_of_Subgroup_Product/Proof_2&oldid=615484"