# Lagrange theorem and subgroup help

1. Aug 18, 2010

### annoymage

1. The problem statement, all variables and given/known data

Let G be group, H<G , K<G, if gcd(lHl,lKl)=1, prove that H$$\bigcap$$K={1}

2. Relevant equations

3. The attempt at a solution

so Lagrange theorem says that lHl l lGl, lKl l lGl,
and of course 1 is inside both H and K, but how when they are coprime, the element are all different except their identity? i cannot see T_T

2. Aug 18, 2010

### Office_Shredder

Staff Emeritus
Re: SubGroup

What must be true about the order of an element of a group

3. Aug 19, 2010

### annoymage

Re: SubGroup

i don't know much about, order of an element, but i noticed the corollary from lagrange theorm, state that order on an element divides order of the group, provided that it is a finite group,

i use oG for order of G for simplicity.

so, gcd(oH,oK)=1 , let h in H, k in K,

so oh l oH and ok l oK, would imply gcd(oh,ok)=1

so their orders are coprime, still cannot see, why their element are different, T_T,

is, cyclic group involved? i still didn't go through cyclic yet

4. Aug 19, 2010

### Office_Shredder

Staff Emeritus
Re: SubGroup

Well, suppose we have an element g which is in both H and K.

What is the order of g? You almost calculated it in your post

5. Aug 19, 2010

### annoymage

Re: SubGroup

yeaaa, i can see now, the order is 1, therefore 1 is the element in both H and K, correct?

6. Aug 19, 2010

Re: SubGroup

No, 1 is in H and K because H and K are subgroups of G.

Are you sure you can assume finiteness here?

Here is a different way to approach. Pick an arbitrary element,k, in K. Now state something about the order of k. Now what happens if you assume k is in H? What is true about all the elements of H?

7. Aug 19, 2010

### annoymage

Re: SubGroup

i mean, 1 is the "only" element in H and K, if it is finite...

hmmmm, so maybe can i divide it into 2 cases? but i do no know what to do if they are infinite, cause i can't use lagrange.

anyway,

let order of k is n, so kn=1, Suppose k also in H, then kn also in H, is that what you mean? T_T and i don't know how to continued

8. Aug 19, 2010

Re: SubGroup

You can get there!!!

Yup, okay, we have k^n = 1. Now what can you tell me, using Lagrange about n and |K|?

Write that down. Now if you assume k is also in H, what does that imply?

9. Aug 19, 2010

### annoymage

Re: SubGroup

yes, but i thought i can only use lagrange for finite group,

and if it is finite, and if k also in H k^n also in H, so n divides both l K l and l H l, will imply gcd(n,n)=1, so n must equal to 1, so order of k is 1, will imply k=1, so 1 is the only element in K and H,
but i did it in the last post, T_T what if it is infinite? i cannot use Lagrange right?

10. Aug 19, 2010