# H normalises [H,K]?

1. Oct 19, 2013

### 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.

Last edited: Oct 19, 2013