Thread: 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 $\displaystyle H$ be a subgroup of a group $\displaystyle G$ and let $\displaystyle g \in G$. Define

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

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

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

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

Give a proof of the statement that is always true

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

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

(i) and (ii) should be straightforward.

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

i) let x,y be in H^g.

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 H^g, so H^g

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

ii) note that (H^g)^k = k(H^g)k^-1 = k(gHg^-1)k^-1 = (kg)H(kg)^-1 = H^kg

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 H^g = ∩_xg (H^g)^x

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

but ∩_xg (H^g)^x = ( ∩_g H^g)^x = N^x, so N is normal.

(to see that for two subgroups H,K, that (H∩K)^g = H^g∩K^g, 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 H^g.

but h is in K as well, so ghg^-1 is in K^g. hence x = ghg^-1 is in both H^g and K^g,

so x is thus in their intersection. on the other hand suppose that x is in H^g and K^g. 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).