Since for all then given you also have So you also have for all i.e. for all
If we take the definition of a normal subgroup to be:
H is normal if for all h in H, g in G.
So it is clear that
However, I want to show that actually , by showing that
I'm just struggling to find a way to prove the final part? Could anyone offer any help?
Definition of normal says for all g in G and h in H. So we have for some h' in H and for all g in G and h in H. This immediately tells us that .
But Holds for all g so in particular we can take and we find:
for some h'' in H and for all h in H. This implies for any h in H and so we conclude that .
Overall and gives equality.
(I only really wrote that our for my benefit, thanks for your help!)
there are various definitions of normal subgroup floating about out there in the ether. one is:
a) gHg-1 = H, for all g in G
b) gHg-1 ⊂ H, for all g in G
and you have just seen that (a) and (b) are equivalent.
a third one is:
c) gH = Hg (left cosets are also right cosets), for all g in G
one could, conceivably, mimic the relationship between (a) and (b) to produce a fourth definition:
d) gH ⊂ Hg, for all g in G.
the thing to see is that all 4 definitions produce the same set of subgroups, for any group G. so which one you use is largely a matter of convenience (for sets, it is often easier to show containment, rather than equality).