1. ## Normal subgroup

If we take the definition of a normal subgroup to be:

H is normal if $ghg^{-1} \in H$ for all h in H, g in G.

So it is clear that $gHg^{-1} \subset H$

However, I want to show that actually $gHg^{-1} = H$, by showing that $H \subset gHg^{-1}$

I'm just struggling to find a way to prove the final part? Could anyone offer any help?

Thanks,

2. ## Re: Normal subgroup

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

3. ## Re: Normal subgroup

Thanks:

Definition of normal says $ghg^{-1} \in H$ for all g in G and h in H. So we have $ghg^{-1} = h'$ for some h' in H and for all g in G and h in H. This immediately tells us that $gHg^{-1} \subset H$.

But $ghg^{-1} = h'$ Holds for all g so in particular we can take $g=g^{-1}$ and we find:

$g^{-1} h g = h''$ for some h'' in H and for all h in H. This implies $h=gh''g^{-1]$ for any h in H and so we conclude that $gHg^{-1} \subset H$.

Overall $H \subset gHg^{-1}$ and $gHg^{-1} \subset H$ gives equality.

(I only really wrote that our for my benefit, thanks for your help!)

4. ## 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).