In Beachy and Blair's 3rd edition of Abstract Algebra the following problem is given:

Let $H$ and $K$ be subgroups of the group $G$, and let $S$ be the set of left cosets of $K$. Define a group action of $H$ on $S$ by $a \cdot (xH) = ax K$ for $a \in H$ and $x \in G$. By considering the orbit of $K$ under this action, show that $|HK| = \frac{|H||K|}{H \cap K}$

First, I think there is a typo, the action of $H$ on $K$-cosets should be $a \cdot (xK) = axK$. Second, I can't seem to obtain the desired outcome from studying the orbit of $K$. I get: $$ \mathcal{O}(K) = \{ hK \ | \ h \in H \} $$ for $hK = h'K$ we need $h'h^{-1} \in K$ where $h,h' \in H$ hence $h'h \in H \cap K$ which gives $h(H \cap K) = h'(H \cap K)$. Thus, it seems to me there are $|H \cap K|$-representatives for each $K$ coset, in short, $$ |\mathcal{O}(K) = \{ hK \ | \ h \in H \}| = \frac{|H|}{|H \cap K|}.$$ On the other hand, the stabilizer of $K$ is $$ G_K = \{ h \in H \ | \ hK = K \} = H \cap K $$ So, the orbit stabilizer theorem gives $$ |H| = |\mathcal{O}(K)||G_K| = \frac{|H|}{|H \cap K|}|H \cap K| = |H|. $$ True, but, no use to the desired outcome describing the size of $HK$.

Question: how do we show $|HK| = \frac{|H||K|}{H \cap K}$ by applying the theory of group action to the given group action of $H$ on $G/K$?

My approach is not productive. Of course, I know this problem can be solved without group actions (see Dummit and Foote page 93, Prop 13 for example). If my orbit or stabilizer is bogus, please let me know the error of my ways. Thanks in advance for your insights.

  • $\begingroup$ It's exercise 7.3.3 to be clear, also, while not stated, clearly $G$ is a finite group. $\endgroup$ – James S. Cook Oct 26 '16 at 14:14

The orbit of $K\in G/K$ under the action of left multiplication by $H$ will be $HK/K$. Even though the set $HK$ is not a group, it is a union of left cosets of $K$ so the notation makes sense. Since the left cosets $hK$ (with $h\in H$) partition $HK$, we have $|HK/K|=|HK|/|K|$. You've already shown that this equals $|H|/|H\cap K|$, so you're good.

If you're not working with finite groups, we can still make sense of $|HK/K|=|H/H\cap K|$ as a true equality. In fact if $K$ is infinite and $H$ finite, then this is a stronger claim than the original one stating $|HK|=|H||K|/|H\cap K|$ (which will be trivial in that case).

Alternatively, we can have $H\times K$ act on $HK$ by $(h,k)x=hxk^{-1}$. The action is transitive and the stabilizer of $e$ will be $\cong H\cap K$, which yields the result as well.

  • $\begingroup$ I don't think you used the group action to show $|HK|$ is connected to the action of $H$ on $G/K$. The step $|HK/K| = |HK|/|K|$ is fine, but, it does not use group actions as desired. Perhaps this was the desired thought process? Just that the orbit of $K$ under $H$ is $HK/K$ hence we can use coset counting (and not the orbit stabilizer theorem) to complete the count? $\endgroup$ – James S. Cook Oct 26 '16 at 14:50
  • $\begingroup$ @JamesS.Cook You did use the orbit-stabilizer theorem to show it equals $|H|/|H\cap K|$. You've already used the group action in the solution. Altogether this is the solution the exercise is pointing towards. $\endgroup$ – arctic tern Oct 26 '16 at 14:53
  • $\begingroup$ So, the connection is that the number of cosets of the stabilizer is the orbit. I was thinking of this as Lagrange's Theorem, but, I should instead think of it as the orbit stabilizer result.... $\endgroup$ – James S. Cook Oct 27 '16 at 16:57

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