Order Formula Under Second Isomorphism Theorem: Does it Hold?

[quasar987]

In the conditions where the second isomorphism theorem applies, one has H/HnK = HK/K so in particular, taking orders (in the finite case), one has the order formula

|HK| = |H|*|K|/|H n K|.

Does anyone know if this formula holds in general, or under lesser hypotheses? Thx.

 

[micromass]

Yes! The formula applies in the case where H and K are finite subgroups. So no normality is required. Of course, the proof will be a lot different as you cannot use the first isomorphism theorem anymore...

Also, note that HK will not be a subgroup anymore! You'll need normality of H or K for this to be a subgroup. But that doesn't mean that the formula doesn't hold anymore...

 

[micromass]

For your convenience, I'll write the proof down:

Let $$X=\{Hg~\vert~g\in G\}$$. Then K acts on X by multiplication (thus Hg=H.g). It follows that

$$HK=\{hk~\vert~h\in H, k\in K\}=\bigcup_{k\in K}{H\cdot k}$$

and $$\{H\cdot k~\vert~k\in K\}$$ is the orbit of the element x:=H. Also, we have that the stabilizer $$K_x=K\cap H$$, thus there are $$|K|/|H\cap K|$$ elements in an orbit. Also, we have |H.k|=|H|. Thus

$$|HK|=|H||K|/|H\cap K|$$

 

[quasar987]

Oh. Very nice, thanks micromass!