H and K normal subgroups, intersection of H and K=<e>

• Dec 2nd 2010, 08:13 AM
kathrynmath
H and K normal subgroups, intersection of H and K=<e>
Let H and K be normal subgroups of G such that H intersect K=<e>. Show that hk=kh for all h in H and k in K.
H and K are normal so ghg^-1 is in H and gkg^-1 is in K.
want to show hk=kh. So basically I'm showing this is abelian.
Can I do ghg^-1=gkg^-1?
ghg^-1g=gkg^-1g
gh=gk
so that works if g=h
• Dec 2nd 2010, 08:41 AM
mulaosmanovicben
Consider the order of the Union of hK where h is an element of H (All the left cosets) This will be equal to the |H|*|K|/|H intersecct K|

I think this will help especially since you know the intersection is of size one.
• Dec 2nd 2010, 08:44 AM
kathrynmath
Since intersection is size one, we are left with |H|*|K|
• Dec 3rd 2010, 12:40 AM
Swlabr
Quote:

Originally Posted by kathrynmath
Let H and K be normal subgroups of G such that H intersect K=<e>. Show that hk=kh for all h in H and k in K.
H and K are normal so ghg^-1 is in H and gkg^-1 is in K.
want to show hk=kh. So basically I'm showing this is abelian.
Can I do ghg^-1=gkg^-1?
ghg^-1g=gkg^-1g
gh=gk
so that works if g=h

Note that,

$\displaystyle khk^{-1} = h_0 \in H$ and $\displaystyle hk^{-1}h^{-1} = k_0 \in K$ as both subgroups are normal.

Then,
$\displaystyle khk^{-1} = h_0h^{-1}h \Rightarrow khk^{-1}h^{-1} = h_0h^{-1} \in H$
and
$\displaystyle hk^{-1}h^{-1} = k^{-1}kk_0 \Rightarrow khk^{-1}h^{-1} = kk_0 \in K$.

Can you finish it from there? You need to apply the fact that the intersection is trivial.