Solving: How H Normalises [H,K]?

[Silversonic]

I'm slightly confused by definition of normalises.

It's shown in my notes that for two subgroups H, K of G we have

[H,K]^{h_1} ≤ [H,K] for all h_1 in H

and then it says

"and so H normalises [H,K]."

It hasn't been fully introduced to me what a subgroup normalising another subgroup means. But a bit of searching around shows H normalises [H,K] means that [H,K]^{h} = [H,K] for all h in H.

Clearly [H,K]^{h} \subseteq [H,K] from above. But showing [H,K]^{h} \supseteq [H,K] I can't seem to work out.

I've tried myself, I've looked around. Even wikipedia explains it;

http://groupprops.subwiki.org/wiki/Subgroup_normalizes_its_commutator_with_any_subset

But I can't understand how [a,h] is a generating set for H, and even if it was how that goes on to help the proof anyway.

Can anyone shed some light my way?

Edit: I apologise for not realising this should be posted in the HW section.
I may have found my own proof if anyone doesn't mind reading it.

To show [H,K] \subseteq [H,K]^{h} amounts to showing that for any a \in [H,K] we have a \in [H,K]^h

This amounts to showing there exists b \in [H,K] such that

a = h^{-1} b h

Equivalent to showing that

b = hah^{-1} = a^{h^{-1}} is an element of [H,K]

By definition

[H,K] = <[h,k] | h \in H, k \in K>

By noting that [h_m,k_m]^{-1} can be written in the form of [h_j,k_j], a can written in the form

a = [h_1 , k_1][h_2,k_2]...[h_n,k_n]

a^{h^{-1}} = [h_1 , k_1]^{h^{-1}}[h_2,k_2]^{h^{-1}}...[h_n,k_n]^{h^{-1}}

It is easily shown that any [h_j, k_j]^{h} for h \in H is an element of [H,K] and hence as [H,K] is a group, a^{h^{-1}} \in [H,K]

There's probably a shorter proof, because the wikipedia page makes it seem like it's about 1 lines worth.

 

[fresh_42]

$A$ normalises $B$ means $aBa^{-1} \subseteq B$ for all $a\in A$. The equality holds, because you can simple rewrite it as $B \subseteq a^{-1}Ba$.