
Normal subgroup
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?
Thanks,

Re: Normal subgroup
Since for all then given you also have So you also have for all i.e. for all

Re: Normal subgroup
Thanks:
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!)

Re: Normal subgroup
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
another is:
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).