Lagrange theorem and subgroup help

[annoymage]

Homework Statement

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

Homework Equations

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

 

[Office_Shredder]

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

 

[annoymage]

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

 

[Office_Shredder]

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

 

[annoymage]

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

 

[icantadd]

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?

 

[annoymage]

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

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,

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?

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

 

[icantadd]

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

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?

 

[annoymage]

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?

 

[icantadd]

what if it is infinite? i cannot use Lagrange right?

Yeah, I was thinking about that...
Does taking the gcd of a group with infinite order make sense?