# Thread: Groups, Subgroups and Normal Subgroups

1. ## Groups, Subgroups and Normal Subgroups

Hi there,

I'm going over some past papers for an exam but with no answers and have come a cropper on this one, any help appreciated!

Let $H$ be a subgroup of a group $G$ and let $g \in G$. Define

$H^g := \{ ghg^{-1} | h \in H \}$

i) Prove that $H^g$ is a subgroup of $G$.

ii) Let $g, k, \in G$. Which of these statements is true

$(H^g)^k = H^{gk} or (H^g)^k = H^{kg}$

Give a proof of the statement that is always true

iii) Set $N := \bigcap_{g \in G} H^g$. Prove that $N \triangleleft G$ and that $N$ is a subgroup of $G$ that is contained in $H$.

Pretty stuck except for ii) $(H^g)^k = H^{kg}$ is obviously the right one (I hope!).

2. ## Re: Groups, subgroups and normal subgroups

(i) and (ii) should be straightforward.

Hint for (iii). Let $G/H$ denote the set of all left cosets of $H$ in $G.$ Each element of $g$ induces a permutation $\pi_g$ of $G/H$ via $\pi_g(aH)=gaH$ for each left coset $aH.$ Let $\Pi=\left\{\pi_g:g\in G\right\}.$ The mapping $f:G\to\Pi$ defined by $f(g)=\pi_g$ is a homomorphism. Prove that $N$ is the kernel of this homomorphism.

3. ## Re: Groups, Subgroups and Normal Subgroups

i) let x,y be in Hg.

this means that x = ghg-1 for some h in H, and y = gh'g-1, for some h' in H.

therefore, xy-1 = (ghg-1)(gh'g-1) = ghh'-1g-1,

and hh'-1 is in H (since H is a subgroup), therefore xy-1 is in Hg, so Hg

is a subgroup of G (by the one-step subgroup test).

ii) note that (Hg)k = k(Hg)k-1 = k(gHg-1)k-1 = (kg)H(kg)-1 = Hkg

iii) first, suppose that a,b are in N. this means that a,b are in every gHg-1, for any g in G. by part (i) above, each gHg-1

is a subgroup, so ab-1 is in every gHg-1, so ab-1 lies in N, thus N is a subgroup.

let x be any element of G, and let n be an element of N. note that N = ∩g Hg = ∩xg (Hg)x

since as g runs over all elements of G, so does xg.

but ∩xg (Hg)x = ( ∩g Hg)x = Nx, so N is normal.

(to see that for two subgroups H,K, that (H∩K)g = Hg∩Kg, suppose that

x is in (H∩K)g. then x = ghg-1, where h is in H∩K. hence h is in H, so ghg-1 is in Hg.

but h is in K as well, so ghg-1 is in Kg. hence x = ghg-1 is in both Hg and Kg,

so x is thus in their intersection. on the other hand suppose that x is in Hg and Kg. so

x = ghg-1 = gkg-1, for some h in H, and k in K.

if ghg-1 = gkg-1, then h = k, so h = k lies in H∩K, so x is in (H∩K)g

the same proof holds for any arbitrary intersection of subgroups of G).