A classmate stated the following today during class: N is a normal subgroup of G, then gng^{-1}= n for all g∈ G and n ∈ N.

I disagreed because although N is a normal subgroup of G, it is not stated that G is abelian. Who do you agree with?

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- December 14th 2012, 03:08 PMbondvalencebondNormal Subgroup Clarification
A classmate stated the following today during class: N is a normal subgroup of G, then gng

^{-1}= n for all g∈ G and n ∈ N.

I disagreed because although N is a normal subgroup of G, it is not stated that G is abelian. Who do you agree with? - December 14th 2012, 03:21 PMHallsofIvyRe: Normal Subgroup Clarification
What do you think is the

**definition**of "normal subgroup"?

The fact that G is or is not abelian is not particularly relevant. Obviously if G**is**abelian then every subset satisfies that formula- and it "just happens" that every subgroup of an Abelian group**is**a normal subgroup. - December 14th 2012, 03:41 PMbondvalencebondRe: Normal Subgroup Clarification
A normal subgroup can be defined in many different ways. One is that G is normal if gng

^{-1}∈ N for all n∈N, g∈G. Another is that gN=Ng for all g∈G. I know G does not have to be abelian to be normal. But how can you rearrange elements if the group is not abelian?

EDIT: Wouldn't you have to rearrange gng^{-1}= (gg^{-1})n = en = n ∈N? But since it's not abelian, we cannot rearrange, right? - December 14th 2012, 04:37 PMbondvalencebondRe: Normal Subgroup Clarification
I'm sorry, I don't want to spam but are my conclusions true or not?

- December 14th 2012, 10:21 PMDevenoRe: Normal Subgroup Clarification
your classmate is wrong. close does not count in mathematics (well.....usually).

if gng^{-1}= n, for all g in G, and n in N, then:

gn = ng, for all g in G, and for every n in N.

this means that every element of n commutes with every element of G, that is: N is contained in Z(G), the center of G.

now every group contained in the center is necessarily normal (can you see why)? but there exist normal subgroups that are not central.

here is an example: S_{3}has the normal subgroup N = {e,(1 2 3), (1 3 2)}. the normality of this subgroup can be shown in various ways:

a) it is of index 2

b) conjugation preserves cycle types, and all cycles of a given type are conjugate

c) it is the sole sylow 3-subgroup of S_{3}

however, this subgroup is NOT central, there is exists g in S_{3}and n in N such that gng^{-1}≠ n:

let g = (1 2) (so g^{-1}= g, as well, because g is of order 2), let n = (1 2 3).

then gng^{-1}= (1 2)(1 2 3)(1 2) = (1 3 2), which does not equal (1 2 3) (but it IS in N). - December 15th 2012, 12:41 AMbondvalencebondRe: Normal Subgroup Clarification
Thanks for clarfying that. Is my logic wrong, though? What I said about being abelian?

- December 15th 2012, 02:05 PMDevenoRe: Normal Subgroup Clarification
if you mean xy = yx iff G is abelian, that is untrue. if G IS abelian, xy = yx, but there are cases where xy = yx but G is not abelian.

for example: in the group D_{4}= {1,r,r^{2},r^{3},s,sr,sr^{2},sr^{3}} where r^{4}= s^{2}= 1, and rs = sr^{3},

it turns out that for the subgroup N = {1,r^{2}}, we have gng^{-1}= n, for ALL g in D_{4}and n in N (that is, 1 and r^{2}commute with everything), but D_{4}is NOT an abelian group.

for xy = yx, we only need for x and y to commute. one of the ways that this can happen is if either x or y is in Z(G), the center of G (the elements that commute with everything).

now if this is gn = ng where g is ANY element in G, and n is in N, we have two possibilities:

1) g is in Z(G), for all g: that is Z(G) = G <---THIS means G is abelian

2) n is in Z(G), for all n: that is N ⊆ Z(G) <---this is the possibility you are overlooking

there is also a third possibility, which is a bit more complicated to explain:

C(N) = G, where C(N) = {x in G: xn = nx, for all n in N} this is called the CENTRALIZER of N in G, and is a subgroup even if N is only a SUBSET of G:

if gn = ng for all n in N (for a specific g), and g'n = ng' for all n in N (for a specific g'), then: (gg')n = g(g'n) = g(ng') = (gn)g' = (ng)g' = n(gg'), so we have closure, and this is all we need for finite groups.

and if g is in C(N), so that gn = ng for all n in N:

g^{-1}n = g^{-1}ne = g^{-1}n(gg^{-1}) = g^{-1}(ng)g^{-1}= g^{-1}(gn)g^{-1}= (g^{-1}g)ng^{-1}= ng^{-1},

so g^{-1}is in C(N) whenever g is.

however, if C(N) = G (that is, G centralizes all of N) then it turns out that this is the same as saying that N is central: C(N) = G iff N ⊆ Z(G).

perhaps all this seems rather complicated, and it SHOULD: you cannot "leap to the conclusion" that you can only "rearrange a product" if we have an abelian group. abelian groups are "nice" this way, but there are "other rungs" on the ladder from:

"very non-abelian" to "abelian". - December 15th 2012, 02:54 PMbondvalencebondRe: Normal Subgroup Clarification
With the given information, would you say it is true or false? See, I feel like because not enough information is given, it is false. If my classmate had been more specific, I would've agreed with her.

- December 15th 2012, 03:12 PMDevenoRe: Normal Subgroup Clarification
what your classmate said is false. your reasoning (that G must be abelian for that to happen) is also false.

- December 15th 2012, 03:16 PMbondvalencebondRe: Normal Subgroup Clarification
Thank you.