Normal Groups

1. Dec 9, 2011

Bachelier

If N is normal in G, is its normal quotient (factor) group G/N normal as well?

And does it imply that any subgroup of G/N which has a form H/N is normal as well?

Is a subgroup of a normal subgrp normal as well. It appears that not always.

Thanks

2. Dec 9, 2011

micromass

Staff Emeritus
G/N normal in what?? In order to be normal you need two groups: N normal in G. So G/N should be normal in what.

Let $N\subseteq H$ with N normal. We have that H is normal in G if and only if H/N is normal in G/N. This is an easy exercise.

No. A normal subgroup of a normal subgroup is called subnormal. Such a group is not always normal. Can you find a counterexample?? (the dihedral group of order 8 should prove interesting)

3. Dec 9, 2011

Bachelier

G/N is normal in G. It just seems hard to prove.
if N is normal in G. for all g in G, gng-1 is in N.
now if H is in G/N then H = aN for some a in g and not in N. now g(an)g-1 is in H only if it is abelian. what then?

4. Dec 9, 2011

micromass

Staff Emeritus
G/N normal in G?? That makes no sense at all. G/N isn't even a subset of G...

5. Dec 9, 2011

Bachelier

isn't G/N the union of left cosets of N?

6. Dec 9, 2011

Bachelier

I'm probably understanding this incorrectly but wolfram says:

For a group G and a normal subgroup N of G , the quotient group of N in G, written G/N and read " G modulo N ", is the set of cosets of N in G.

http://mathworld.wolfram.com/QuotientGroup.html

7. Dec 9, 2011

Bachelier

Again: The elements of the quotient group G/N are subsets of G

http://people.brandeis.edu/~igusa/Math30/quotient.pdf [Broken]

Last edited by a moderator: May 5, 2017
8. Dec 9, 2011

micromass

Staff Emeritus
No, it's the set of left cosets of N. Not the union.

Elements of G/N are subsets of G (not elements of G). So G/N can't be a subset of G.

9. Dec 9, 2011

micromass

Staff Emeritus
Yes. That doesn't mean that G/N is a subset of G.

Last edited by a moderator: May 5, 2017
10. Dec 9, 2011

Bachelier

So we know G/N is a group, but then it isn't a subgroup of anything.

How do you imagine G/N like in space (as a group)?

11. Dec 9, 2011

Bachelier

so I guess every time I think of quotient groups I need to do so in terms of mappings

12. Dec 9, 2011

Bachelier

in G/N

consider:

f: G ---> G/N ​

g---> gN​

f is onto homomorphism.

now image of Hunder this map is H/N

I don't know if I can use the 3rd isomorphism thm b/c N is not given to us to be normal. But at least, H/N the image of H is normal because f(aha-1) is in the image(H) and since homomorphism then f(a)*f(h)*f(a)-1 in H/N.

Last edited: Dec 9, 2011
13. Dec 9, 2011

Bachelier

please let me know if this is correct.

Thanks a lot, without you I wouldn't have figured this out.

14. Dec 9, 2011

Bachelier

let a, b in D4 s.t. |b|=2 and |a|= 4

then K= {e, a2, b, a2b} is a normal subgroup in D4

but H = {e, b} is a subgroup in K however not normal.

15. Dec 10, 2011

Deveno

more importantly, H is normal in K (since K is abelian, every subgroup of it is normal in it), and not normal in G.

normality isn't "transitive".

there are two (equivalent) ways of thinking of a factor group (quotient group):

1) as a homomorphic image of the original group.
2) as a group induced by partitioning by a normal subgroup.

way number 2 is what involves cosets, which are the equivalence classes under the partition. we require normality of the subgroup, so that we can multiply cosets:

NaNb = Nab for all a,b in G iff xN = Nx for all x in G.

if we try to form G/H with H some non-normal subgroup, the sets Hab and HaHb aren't the same for at least one pair (a,b) in GxG. the set of cosets can still be formed, we just don't get a group out of it. try this with D3 or D4 and some non-normal subgroup.

the equivalence of 1 and 2 is what the first isomorphism theorem is all about.

16. Dec 12, 2011

someriraq

thanks all ...