
Normal subgroup
If we take the definition of a normal subgroup to be:
H is normal if $\displaystyle ghg^{1} \in H $ for all h in H, g in G.
So it is clear that $\displaystyle gHg^{1} \subset H $
However, I want to show that actually $\displaystyle gHg^{1} = H $, by showing that $\displaystyle H \subset gHg^{1} $
I'm just struggling to find a way to prove the final part? Could anyone offer any help?
Thanks,

Re: Normal subgroup
Since $\displaystyle ghg^{1}\in H$ for all $\displaystyle g\in G,$ then given $\displaystyle g\in G$ you also have $\displaystyle g^{1}h\left(g^{1}\right)^{1}\in H.$ So you also have $\displaystyle g^{1}Hg\subseteq H$ for all $\displaystyle g\in G,$ i.e. $\displaystyle H\subseteq gHg^{1}$ for all $\displaystyle g\in G.$

Re: Normal subgroup
Thanks:
Definition of normal says $\displaystyle ghg^{1} \in H $ for all g in G and h in H. So we have $\displaystyle ghg^{1} = h' $ for some h' in H and for all g in G and h in H. This immediately tells us that $\displaystyle gHg^{1} \subset H $.
But $\displaystyle ghg^{1} = h' $ Holds for all g so in particular we can take $\displaystyle g=g^{1}$ and we find:
$\displaystyle g^{1} h g = h'' $ for some h'' in H and for all h in H. This implies $\displaystyle h=gh''g^{1] $ for any h in H and so we conclude that $\displaystyle gHg^{1} \subset H$.
Overall $\displaystyle H \subset gHg^{1} $ and $\displaystyle gHg^{1} \subset H $ 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).