# the order of two elements in group

• Nov 27th 2012, 07:16 AM
cummings123321
the order of two elements in group
Let G be an abelian group and let g,h in G, assume that |g|,|h| (the order of g and h) are both finite,with hcf(|g|,|h|)=1.then Prove that |g+h|=|g|*|h|
i dont know how to relate the order of g,h to the order of |g+h| could somone give me any hints ??
• Nov 27th 2012, 07:33 AM
ModusPonens
Re: the order of two elements in group
Start summing g+h with g+h. you'll get m(g+h), where m is an integer. since hcf(|g|,|h|)=1, the intersection of <g> and <h> is the {0}. thus, m(g+h)=mg+mh is always different from 0 unless both mg=0 and mh=0. That happens when m=lcm(|g|,|h|)=|g|*|h|.
• Nov 27th 2012, 07:37 AM
cummings123321
Re: the order of two elements in group
why hcf(|g|,|h|)=1 implies the intersection of <g> and <h> is the {0}????
• Nov 27th 2012, 07:43 AM
Re: the order of two elements in group
Suppose it isn't true, let t be an element of the intersection. t=ag=bh. we have t*|t|=(ag)*|t|=(bh)*|t|=0, the orders of g and h are a|t| and b|t|, which is a contradiction cuz they are supposed to have hcf=1
• Nov 27th 2012, 07:56 AM
emakarov
Re: the order of two elements in group
You need the theorem that if ng = 0, then |g| divides n. Using it, show that |g + h| divides |g| * |h|. Next, you need to prove that both |g| and |h| divide |g + h|. Since hcf(|g|, |h|) = 1, this would imply that |g| * |h| divides |g + h| and so |g + h| = |g| * |h|.

To prove that |g| divides |g + h|, consider |g + h|(g + h) = 0 and multiply both sides by the number (not a group element) |h|. Then use the fact above and (a generalization) of the Euclid's lemma.
• Nov 27th 2012, 07:58 AM
emakarov
Re: the order of two elements in group
Quote:

Suppose it isn't true, let t be an element of the intersection. t=ag=bh. we have t*|t|=(ag)*|t|=(bh)*|t|=0, the orders of g and h are a|t| and b|t|

The last fact does not follow.
• Nov 27th 2012, 09:46 AM
Deveno
Re: the order of two elements in group
let |g| = k, |h| = m.

consider (km)(g+h) = (km)g + (km)h = (mk)g + (km)h (since km = mk in the the ring of integers)

= m(kg) + k(mh) = m(0) + k(0) = 0 + 0 = 0 (we are using the fact that an abelian group has a natural interpretation as a Z-module (the "laws of exponents" with "powers" meaning "repeated sums")).

since (km)(g + h) = 0, |g + h| divides km.

now suppose we write |g + h| = t.

this means that t(g +h) = tg + th = 0.

thus -th = tg, which shows that -th is in both <h> and <g>. so we'll be done if we can show that <g> ∩ <h> = {0}

(because this will show that -th = tg = 0, and thus th = -0 = 0, so k divides t, and m divides t).

by lagrange, |<g> ∩ <h>| divides <g> = k, and <h> = m. but gcd(k,m) = 1, so |<g> ∩ <h>| = 1, and thus must be the trivial subgroup.
• Nov 27th 2012, 01:29 PM
cummings123321
Re: the order of two elements in group
thank you i see , many thanks 2 u!!!
• Nov 27th 2012, 01:34 PM
emakarov
Re: the order of two elements in group
Quote:

Originally Posted by cummings123321
why hcf(|g|, |h|) = 1, this would imply that |g| * |h| divides |g + h|????

You first prove that |g| and |h| divide |g + h|, as describe in post #5.
• Nov 27th 2012, 01:52 PM
cummings123321
Re: the order of two elements in group
yes,i get it now ,thanks a lot