# Herstein: Normal subgroup proof.

• Oct 15th 2011, 11:23 AM
ThatPinkSock
Herstein: Normal subgroup proof.
Herstein: Page 74 #18 (sect 2.5)

If H is any subgroup of G and N = ∩(a∈G) a^-1Ha, prove that N◅G.

I am extremely lost. So lost, it hurts.
• Oct 15th 2011, 01:47 PM
Deveno
Re: Herstein: Normal subgroup proof.
first of all, for any a in G, and any subgroup H of G, a^-1Ha is a subgroup of G (possibly equal to H, possibly not).

for any two such groups, their intersection (a^-1Ha)∩(b^-1Hb) is also a subgroup of G.

if we take the intersection of ALL such subgroups (over every a in G), we still get a subgroup (notice this is non-empty because {e} is in every one).

our task is to show that that intersection is normal, so we want to show that for any g in G, g^-1Ng is contained in N.

so let's take a "typical" element g^-1ng in g^-1Ng, so n is a "typical" element in N. this means that n is in every subgroup

a^-1Ha, for any a in G. therefore g^-1ng is in every subgroup g^-1(a^-1Ha)g = (ag)^-1H(ag) for any a in G.

if we knew the set of subgroups {(ag)^-1H(ag):a in G} was the same set of subgroups {a^-1Ha : a in G}, we'd be done,

because that would show g^-1ng was in N. this is the same thing as saying that the map a ---> ag is a bijection. is this true?

suppose ag = ag'. then a^-1(ag) = a^-1(ag'), so g = g'. so the map a-->ag is injective.

suppose h is any element of G. can we find some a in G such that ag = h? sure, let a = hg^-1. this shows a-->ga is surjective,

and thus a bijection, hence, as a runs through G, ag also runs through G, so the intersection ∩(a∈G) (ag)^-1H(ag) = ∩(a∈G) a^-1Ha,

and g^-1ng is in the first intersection, so is in the second intersection, which is what N is. so N is normal

(this subgroup is called the normal core of H).
• Oct 16th 2011, 03:07 PM
ThatPinkSock
Re: Herstein: Normal subgroup proof.
Quote:

Originally Posted by Deveno
first of all, for any a in G, and any subgroup H of G, a^-1Ha is a subgroup of G (possibly equal to H, possibly not).

for any two such groups, their intersection (a^-1Ha)∩(b^-1Hb) is also a subgroup of G.

if we take the intersection of ALL such subgroups (over every a in G), we still get a subgroup (notice this is non-empty because {e} is in every one).

our task is to show that that intersection is normal, so we want to show that for any g in G, g^-1Ng is contained in N.

so let's take a "typical" element g^-1ng in g^-1Ng, so n is a "typical" element in N. this means that n is in every subgroup

a^-1Ha, for any a in G. therefore g^-1ng is in every subgroup g^-1(a^-1Ha)g = (ag)^-1H(ag) for any a in G.

if we knew the set of subgroups {(ag)^-1H(ag):a in G} was the same set of subgroups {a^-1Ha : a in G}, we'd be done,

because that would show g^-1ng was in N. this is the same thing as saying that the map a ---> ag is a bijection. is this true?

suppose ag = ag'. then a^-1(ag) = a^-1(ag'), so g = g'. so the map a-->ag is injective.

suppose h is any element of G. can we find some a in G such that ag = h? sure, let a = hg^-1. this shows a-->ga is surjective,

and thus a bijection, hence, as a runs through G, ag also runs through G, so the intersection ∩(a∈G) (ag)^-1H(ag) = ∩(a∈G) a^-1Ha,

and g^-1ng is in the first intersection, so is in the second intersection, which is what N is. so N is normal

(this subgroup is called the normal core of H).

Thank you so much! This is great. (Nod)