Question

hey could someone please explain how normal subgroups work? with some good examples if you dont mind

Answer 1

Yep, absolutely. A normal subgroup is a subgroup N of a group G such that every element of N, when conjugated by an element of G, goes back to another element of N. Conjugation meaning multiplication of the following type: \(g^{-1}ng\). This is a pretty abstract definition that isn't super easy to understand right away, so some examples are definitely in order.

First example: you've probably already heard that every subgroup of an abelian group is normal. This becomes pretty clear when you look at the definition. If you want \(g^{-1}Ng\to N\), and the subgroup is abelian, you can just move the \(g\) through by commuting the multiplication, giving you \(g^{-1}gN\), which cancels to N.

Second example: The kernel of a homomorphism is always a normal subgroup (and vice versa, every normal subgroup is the kernel of some homomorphism). Here are a couple of classical examples. One is the kernel of the determinant homomorphism on GLn(R). That kernel is the group SLn(R), which is normal in GLn(R). Another example is the sign homomorphism of Sn. The kernel of that homomorphism is An, which is a normal subgroup of Sn.

Hopefully that helps a bit, let me know if you have other specific questions about the concept.

Answer 2

hmm yeah it does i am trying to do this one practice problem which i dont quite get at all would you mind helping me out with it like yeah i know the definition and your explanation filled in some blanks i had

Answer 3

so basically if that group is abelian then it has to be normal?

Answer 4

Yeah, every subgroup of an abelian group is always normal, because you can just commute the multiplication.

Answer 5

but how would we find out if one is abelian

Answer 6

Computation, mostly.

Answer 7

Is there a direct method of finding if a group is abelian besides computing elements?

Answer 8

In general, if you conjugate in an abelian group, it doesn't do anything. Proof:\[
g,h\in G\text{ (abelian)}\\
gh=hg \text{ by abelian property}\\
ghg^{-1}=hgg^{-1}\text{ multiply on the right by g inverse}\\
ghg^{-1}=h \text{ cancel}
\]So, conjugation in an abelian group sends every element to itself.

As far as finding out whether a group is abelian, generally you can just look the group up, or think about some examples. If a group is non-abelian it's usually pretty easy to come up with an example or two of elements that don't commute with each other.

Answer 9

Important examples to know are that Sn for n>2 is always non-abelian, Z/n is always abelian, GLnR and SLnR are non-abelian.

Answer 10

could we try one example that i got

Answer 11

Except for when n=1, then GLnR and SLnR are abelian.

Answer 12

Yeah, go ahead and post it.

Answer 13

1) The dihedral group of order 12 is D6 = {e, r, r^2, r^3, r^4, r^5, s, rs, (r^2)s, (r^3)s, (r^4)s, (r^5)s } where |r| = 6, |s| = 2 and sr = (r^5)s
a) Let H = <r^2> and show that H is normal to D6( the sideway triangle symbol).
b) Construct the Caylet table of D6/H. Is this group abelian? Is it cyclic?

Answer 14

D6 is non-abelian. In general, the dihedral groups are non-abelian. They're abelian for n<3, but those groups aren't usually referred to as dihedral.

Answer 15

So, with that in mind, to show normal you have to take a different approach. First step, I would compute some conjugations of that group to get an idea for what it looks like.

Answer 16

hmmm

Answer 17

Oh, by the way, a good way to know that D6 is non-abelian even if you didn't already know it is that they told you in the question. When they said \(sr=r^5s\), that means it's non-abelian because \(sr\ne rs\).

Answer 18

oo yeah thats truee but how do we show that H is normal to D6? well clearly it is nt because it is non abelian

Answer 19

is that sufficient enough for 1a ? :s

Answer 20

H is normal to D6. Abelian implies normal, but the converse isn't true. So, what you need to do is try conjugating elements of H by elements of D6, and see what happens.

Answer 21

what do you mean by that could we do one element from D6? and try it

Answer 22

Yeah, take elements from D6 and conjugate H by them, and see what you get as a result

Answer 23

Start with r as an example. Conjugate each element of H by r.

Answer 24

im sorry @nbouscal but i still dont get it

Answer 25

Okay, well, first off, can you explicitly list the elements of H?

Answer 26

i have done using the left and right cosets with permutation groups but not with like r and r^2 and stuff

Answer 27

H = (r^2)

Answer 28

H is the subgroup generated by r^2, so it's more than just that one element.

Answer 29

Remember what is required to be a subgroup: it must be closed under the multiplication and closed under inverses, and it must contain the identity.

Answer 30

alright so what i think it is

eH
rH
r^2H .... ?

Answer 31

eH = (r^2)
rH = (r^3)
r^2H=(r^4) ? and i do this for all the elements or am i totally wrong

Answer 32

No, not at all that. We're still just looking for what the elements of H are to begin with.

Answer 33

H is a subgroup, so it has to fulfill all the requirements to be a subgroup. It's generated by r^2, so that means it has all powers of r^2 in it.

Answer 34

First step is it has to have e, because every subgroup has to have the identity. Then we can also look and see, what is the inverse of r^2? Because it has to be closed under inverses.

Answer 35

but doesnt for one group to be normal the left coset has to equal to the right coset?

Answer 36

Yeah, we're not that far yet though, we still have to find the subgroup itself to begin with

Answer 37

OO are we showing first if it is a group by showing associativity inverse and identity?

Answer 38

We're just finding out what the group is. What are the elements of H?

Answer 39

We have to be able to list out the elements of H explicitly, so that we know what we're multiplying by when we figure out cosets

Answer 40

what do you mean by writing it explicitly i dont get that

Answer 41

I mean, what are the elements of the group? We have listed the 12 elements of D6, we need to list the elements of H.

Answer 42

but the element of H is given to us which is (r^2) is it not

Answer 43

No. If H was just r^2, it wouldn't be a subgroup! Subgroups have to have the identity, and have to be closed under multiplication and inverses. Just {r^2} is not a subgroup!

What was given to us is <r^2>, which means the subgroup generated by r^2.

Answer 44

Perhaps you ought to review what generator terminology means, what it means to have a group generated by an element. Cyclic subgroups and cyclic groups.

Answer 45

gosh this is confusing i will try learning more of this stuff and post questions later on then

Answer 46

Okay. To answer the question for you, H is {e, r^2, r^4}. Definitely review generators, subgroups, cyclic subgroups, etc.

Answer 47

@hamza_b23, did my response to your question about injective, surjective, and bijective help? Just curious.

Answer 48

yeah it did help a lot thank you