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I'm not being able to understand a statement in the proof of the following theorem:

Theorem: If $S,T$ are subgroups of a finite group $G$ then

$$|ST||S\cap T|=|S||T|$$

He starts by defining $\varphi:S \times T \to ST$ by $\varphi (s,t)=st$. He proceeds by stating that since $\varphi$ is a surjection, it suffices to show that if $x \in ST$ then $|\varphi^{-1}(x)|=|S\cap T|$. With no further explanation he demonstrates de later statement and concludes the demonstration. So Im having trouble in understanding why $\varphi$ being a surjection implies that it is enough to show that $|\varphi^{-1}(x)|=|S\cap T|$ to conclude the proof.

Bill Dubuque
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Joaquin Liniado
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2 Answers2

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The author simply uses the shepherd's principle:

Shepherd's principle (1st version)

Let $\mathcal P$ be a partition of a finite set $E$, and suppose each $X\in\mathcal P$ has the same cardinality $n$. Then $\;\lvert E\rvert=n\lvert\mathcal P\rvert$.

Shepherd's principle (2nd version)

Let $f\colon E\longrightarrow F$ a surjective map between finite sets. If each $f^{-1}(x)$ $\;(x\in E)$ has the same cardinality $n$, then $\;\lvert E\rvert=n\lvert F\rvert$.

Recall that the fibers of a map partition the domain. Above is the special case when the fibers have constant size.

Bill Dubuque
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Bernard
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  • Which is? It´s not even in Wikipedia.. – Joaquin Liniado Sep 29 '16 at 22:14
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    It's a metaphoric name. It says that, to count sheep in his flock, a shepherd counts legs and divides the result by four. – Bernard Sep 29 '16 at 22:22
  • Mhm.. and how does the shepard's principle applies to this precise situation? – Joaquin Liniado Sep 29 '16 at 22:24
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    Each inverse image by φ has the same number of elements, just as, in the mapping from the set of legs onto the set of sheep, each ‘inverse image of a sheep’ has 4 elements. It has nothing to do with group theory, only with combinatorics. – Bernard Sep 29 '16 at 22:28
  • Yes. I see. Still don't understand the logic behind this statement and the product formula. – Joaquin Liniado Sep 29 '16 at 22:37
  • You have a surjection from S×T onto ST. |S×T| = |S| |T| and each inverse image has cardinality |S$\cap$T|, isn't it clear these data yield the result? – Bernard Sep 29 '16 at 22:42
  • No! That is precisely what Im asking in this question! – Joaquin Liniado Sep 29 '16 at 22:44
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    |S×T| is equal to the sum of the number of elements in each subset member of the partition. These subsets all have |S$\cap$T| elements, so you can factor this number. The other factor is the number of subsets in the partition, which is the number of elements in ST, by definition. – Bernard Sep 29 '16 at 22:58
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Maybe a different point of view can help your understanding. The map $\varphi\colon S\times T\to ST$ defined by $\varphi(s,t)=st$ is surjective. Define an equivalence relation on $S\times T$ by declaring $$ (s,t)\sim(s',t') \text{ if and only if }\varphi(s,t)=\varphi(s',t') $$ (that is, $st=s't'$). Then $\varphi$ induces a bijection $S\times T/{\sim}\to ST$, so the number of equivalence classes is $|ST|$.

If we prove that all equivalence classes with respect to $\sim$ have the same number of elements $k$, we'll know that $|ST|=|S\times T|/k$, because $S\times T$ is partitioned in classes with the same number of elements $k$.

Now the equivalence class of $(s,t)$ is precisely $\varphi^{-1}(st)$. When we have proved that, for each $x\in ST$ we have $|\varphi^{-1}(x)|=|S\cap T|$ we have precisely determined we're in the hoped for situation and that $k=|S\cap T|$.

The idea is the same as the common proof of Lagrange's theorem: a subgroup $H$ of $G$ defines an equivalence relation $x\sim y$ if and only if $x^{-1}y\in H$. The equivalence classes have the same cardinality as $H$, so $|G|$ is $|H|$ times the number of equivalence classes. In the present case the number of equivalence classes is $|ST|$ and their common cardinality is $|S\cap T|$.

The proof can be done as follows: let $x\in ST$. Fix $(s_0,t_0)$ such that $x=s_0t_0$; for $(s,t)\in\varphi^{-1}(x)$, consider that $st=s_0t_0$, so $s_0^{-1}s=t_0t^{-1}\in S\cap T$; then define $$ f\colon \varphi^{-1}(x)\to S\cap T $$ by $f(s,t)=s_0^{-1}s=t_0t^{-1}$ and prove it is a bijection.

egreg
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