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Attempt so far is that I've tried using the Lagrange theorem but i'm having problems in showing the first result about the double cosets. In regards to the last three statements, I'm completely stumped. Any help will be appreciated. thanks.

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    $\begingroup$ Hint: what is the double coset $HeK$? Can you think of a group of order $6$ with a double coset of cardinality $4$? Investigate some double cosets of subgroups of $S_3$ of order $2$. $\endgroup$ Aug 5 '15 at 2:19
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In general, whenever you have to show that every element of a set lie in a given family of its subsets and those subsets are mutually disjoint this indicates that you are dealing with equivalence relations/partitions.

So what is the relation involved? Let $a,b \in G$, define a relation from $G \longrightarrow G$ as follows: $$a \equiv b \qquad \iff \qquad \exists g \in G \, \text{ such that } a \in HgK \text{ and } b \in HgK.$$ This is same as saying $$a \equiv b \qquad \iff \qquad \exists h \in H, k \in K \text{ such that } a=hbk.$$ Try proving that this is an equivalence relation.

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  • $\begingroup$ sure thats a nice approach, I can do that, any insight for the last part? $\endgroup$
    – Raul
    Aug 5 '15 at 2:18
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If $A\to B$ is an onto map and $\Gamma$ a set partition of $B$, then the preimages of cells of $\Gamma$ form a set partition of the domain $A$. Consider the map $G\to G/K$ and partition $G/K$ into $H$-orbits; this proves the set of double cosets $H\backslash G/K=\{HgK:g\in G\}$ partitions $G$. Alternatively one can have the product $H\times K$ act on $G$ via $(h,k)g:=hgk^{-1}$ and then $G$ mod $H\times K$ is in fact the double coset space $H\setminus G/K$.

(1) The simplest double coset is $HK$. Its size is $|H||K|/|H\cap K|$. What if $|H\cap K|=1$ and the product $|H||K|$ has too many of $|G|$s prime factors (counted with multiplicity) to divide $|G|$?

The easiest way to arrange for that is if $H,K$ are distinct cyclic groups of prime order $p$ (hence trivially intersecting) and $p^2\nmid |G|$. The smallest prime is $p=2$, so let's use that. We need a nonabelian group; the smallest is $S_3$. What happens with distinct cyclic subgroups of order $2$?

(This is in fact the example hint given in David Wheeler's comment.)

(2) Here's a trick: $|HgK|=|HgKg^{-1}|$. Using the previous counting formula again, the question reduces to if $|H\cap gKg^{-1}|$ depends on $g$. What if $H,K$ are conjugate but unequal?

(3) Might as well check the example in (1) just in case you get lucky. In fact, one could just as easily invoke the example of (1) in problem (2) as well!


As $HK=\bigcup_{h\in H}hK$ is a union of left cosets of $K$, we can form the coset space $HK/K$ on which the group $H$ acts transitively from the left. The stabilizer of $K$ is $H\cap K$, so there is an equivariant bijection $HK/K\to H/(H\cap K)$ given by $hK\mapsto h(H\cap K)$. (Orbit-stabilizer!) This is what justifies the numerical equality between the two, even without restrictions on finiteness.

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Two years late to the party, but this came up in my homework, so here you go.

One intuitive way to prove that a certain relation determines a partition on a set is to prove that the relation in question is an equivalence relation. Equivalence relations defined on the elements of a set partition the set.

Let $G$ be a group and let $H$ and $K$ be subgroups of $G$. Let $a, b \in G$ and define the following congruence relation: $$a \equiv b\ \ \text{if}\ \ b = hak \ \text{ for some } h \in H \text{ and some } k \in K. $$

To show that this relation is an equivalence relation, we need to verify the defining properties. Assume all elements are elements of $G.$

  1. Transitivity:

Suppose $a \equiv b $ and $b \equiv c,$ so $$b = hak \text{ and } c = h'bk'$$ for some $h, h' \in H$ and some $k, k' \in K$.

Substituting for $b$ yields $$c = h'hakk'.$$

Since $H$ and $K$ are groups, $h'h \in H$ and $kk' \in K$, so $a \equiv c.$

  1. Symmetry

Suppose $a \equiv b$, so $$b = hak \text{ for some } h \in H, k \in K,$$ then $$a = h^{-1}bk^{-1},$$ where $h^{-1} \in H$ and $k^{-1} \in K,$ so $b \equiv a$.

  1. Reflexivity

Finally, $$a = 1a1,$$ where $1 \in H$ and $1 \in K,$ so $a \equiv a.$

As you might have noticed, each of these follows from a property of a subgroup.

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    $\begingroup$ glad to have you at the party $\endgroup$
    – Raul
    Nov 13 '17 at 1:46
  • $\begingroup$ @Raul What did you end up doing with your math undergrad? $\endgroup$ Nov 13 '17 at 9:18

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