Question

Let g be a group of order ab where g has two subgroups h and k of order a and b respectively. Show that if h k = (1) then g = hk.

Answer 1

Answer:

Hello.

I hope this helps you.

The key is to notice that since |h| = a and |k| = b, there are ab products of the form uv with u ∈ h and v ∈ k.  IF all these products give different vallues, then since there are only ab values in |g| in total, this will mean that g = hk.

All that's left then is to show is:

• if uv = xy, with u,x ∈ h and v, y ∈ k, then u = x and v = y,

because this will mean that whenever we take different pairs u, v, we get different results.

Suppose uv = xy.  Let's write x' for the inverse of x and v' for the inverse of v.  Then

uv = xy

=> x'uv = y

=> x'u = yv'

Since u, x ∈ h, we have u, x' ∈ h and so x'u ∈ h.

Since v, y ∈ k, we have v', y ∈ k and so yv' ∈ k.

So x'u = yv' is a value in both h and k.  Since h ∩ k = { 1 }, it follows that

x'u = yv' = 1

=> x'u = 1  and  yv' = 1

=> u = x  and  y = v.