- #1

- 130

- 1

I'm slightly confused by definition of normalises.

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

[itex] [H,K]^{h_1} ≤ [H,K] [/itex] for all [itex] h_1 [/itex] in [itex] H [/itex]

and then it says

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

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 [itex] [H,K]^{h} = [H,K] [/itex] for all h in H.

Clearly [itex] [H,K]^{h} \subseteq [H,K] [/itex] from above. But showing [itex] [H,K]^{h} \supseteq [H,K] [/itex] 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.

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

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

[itex] a = h^{-1} b h [/itex]

Equivalent to showing that

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

By definition

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

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

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

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

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

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

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

[itex] [H,K]^{h_1} ≤ [H,K] [/itex] for all [itex] h_1 [/itex] in [itex] H [/itex]

and then it says

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

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 [itex] [H,K]^{h} = [H,K] [/itex] for all h in H.

Clearly [itex] [H,K]^{h} \subseteq [H,K] [/itex] from above. But showing [itex] [H,K]^{h} \supseteq [H,K] [/itex] 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 [itex] [H,K] \subseteq [H,K]^{h} [/itex] amounts to showing that for any [itex] a \in [H,K] [/itex] we have [itex] a \in [H,K]^h [/itex]

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

[itex] a = h^{-1} b h [/itex]

Equivalent to showing that

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

By definition

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

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

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

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

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

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

Last edited: