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!

 

[blue_raver22]

I would say that the order formula under the second isomorphism theorem holds under the specific conditions where the theorem applies. This means that the groups H and K must be subgroups of a larger group G, and HnK must be a normal subgroup of G. In this case, the formula holds in both the finite and infinite cases.

However, it is important to note that this formula may not hold under lesser hypotheses. For example, if H and K are not subgroups of a larger group G, or if HnK is not a normal subgroup of G, then the formula may not hold. It is always important to carefully consider the conditions and assumptions when applying mathematical theorems.