Question

G is a group
H is a normal subgroup of G
prove: if (G:H)= p where p is prime, then the order of every element a in G not in H is a multiple of p.

Answer 1

hint: for each a in G, order of the element Ha in G/h is a divisor of the order of a in G.

Answer 2

I don't even know where to start...

Answer 3

Well, first you could notice that \(G/H\) must be cyclic since it has prime order.

Answer 4

I actually did think about that..but don't know what to do with it.

Answer 5

Well, this might be related to the problem you posted earlier.

Answer 6

Do we have to assume that G is finite?

Answer 7

Well, let \(r\) be the order of \(g\in G\), \(g\notin H\). We want to show that \(p|r\). Consider the order \(s\) of \(Hg\) in \(G/H\). We know from that theorem that \(s|r\). But since \(G/H\) has prime order, that means that any element must have order 1 or order \(p\). But it only has order 1 if \(g\in H\). Since we assumed this to not be the case, \(s=p\). Thus, \(p|r\).

Answer 8

we were not told G is finite @terenzreignz

Answer 9

I don't think you need \(G\) to be finite. You just need \(G/H\) to be finite and have prime order.

Answer 10

Well, what if G is the set of integers (under usual addition)
And H is the set of even integers.

H is a cyclic (and therefore abelian, and therefore normal) subgroup of G.

there are two elements of G/H
H and 1+H
2 is prime.

But no element in G would have finite order.

Answer 11

hmm

Answer 12

Well, except 0 anyway

Answer 13

Fair enough. Perhaps it is implied somewhere, or perhaps it was stated at the beginning of the section or chapter or problem set.

But in any case, the proof I provided does not rely on the finiteness/infiniteness of the group.

Answer 14

It does not state it, and it normally does at the begging of the question set. Having said that, there are mistakes in the book.

but, how can the proof be right, when that is a counter example? Not that I think its wrong....

Answer 15

Other than assuming \(g\) has finite order.

Although (I would have to think about this), it may be possible that \(g\in G\), \(g\notin H\) may imply finite order if \(G/H\) has prime order.

Answer 16

If that is true, then the statement of the problem is fine.

Answer 17

But then the counter example?

Answer 18

The counter example doesn't work. How is the set of even integers cyclic?

Answer 19

Generated by 2

Answer 20

\[\large set \ of \ even \ integers = <2>\]

Answer 21

much like how
\[\large \mathbb{Z} = <1>\]

Answer 22

How does that generate -2?

Answer 23

2-2-2

Answer 24

It generates the positive integers, but not all the even integers.

\(2a=-2\) for what positive \(a\)?

Answer 25

a doesn't have to be positive, see

Answer 26

If you start with one generator, it does.

Answer 27

a just has to be an integer

Answer 28

\(\{2,-2\}\) generate all the even integers. But since we use addition as our group operation, we need \(2+2+...=-2\) to let 2 generate the even integers.

Answer 29

I thought that <a> implies that <a> is generated by a and a^(-1)

Answer 30

otherwise, not even \(\Large \mathbb{Z}\) is cyclic.

Answer 31

But we know for a fact that the set of integers is cyclic...

Answer 32

Hmmm... Hold on. Let me review my definitions.

Answer 33

Here, satisfy yourself :D
http://en.wikibooks.org/wiki/Abstract_Algebra/Group_Theory/Cyclic_groups/Definition_of_a_Cyclic_Group

Answer 34

it doesn't say n has to be a positive integer.

Answer 35

I rescind my previous few statements. It seems you are correct about this.

Answer 36

So, my counterexample does work, then? :D

Answer 37

it says <g> = {g^n|n in Z}

Answer 38

but the proof is solid....lol so something is up.

Answer 39

Anyway, we didn't need the set of evens to be cyclic... just abelian, so that it's also a normal subgroup

Answer 40

True enough. But then could you make the argument that since 2 (kind of in a really strange way) divides infinity it's still true.

But then again, 2 doesn't *really* divide infinity, so....

Answer 41

I think I see it... the part that says...
We know from that theorem that s|r
this kinda only applies if G was finite to begin with

Answer 42

that's what I was typing. if order is infinite then infinity = p*k for some k:)

Answer 43

or specifically, that the order "r" of g (which is in G but not in H) is a finite numer.

Answer 44

but that's not true @zzr0ck3r since \(k\notin\mathbb{Z}\), it doesn't technically divide infinity.

My opinion is that the problem meant for you to assume \(G\) was finite.

Answer 45

Or it was mentioned somewhere unseen...
Sorry guys, I have to go to class now... this was really fun ^.^

Answer 46

cya later terenz

Answer 47

yeah because even the hint is from question 1 of the same set, with the same assumptions

Answer 48

pz dude ty for your help terenz

Answer 49

@kinggeorge

http://books.google.com/books?id=qviuEUavE4IC&pg=PA154&lpg=PA154&dq=E.+Order+of+Elements+in+Quotient+groups+pinter&source=bl&ots=bFZV4RC9C8&sig=nbYGuE3JtGdMJfHQ-QmqfdvBF14&hl=en&sa=X&ei=2MSuUbfvCOiGjAKftYDYBw&ved=0CCoQ6AEwAA#v=onepage&q=E.%20Order%20of%20Elements%20in%20Quotient%20groups%20pinter&f=false

Answer 50

Part E

Answer 51

Hmm. I definitely can't find anywhere that suggests that \(G\) be finite. I'm getting the feeling that it may be an error.

Answer 52

same here...it must be. Can you prove the hint without G being finite?

Answer 53

that's what im trying to do now...

Answer 54

No, if \(a\in G\) has infinite order, then it's basically game over.