# Thread: Proof of intersect between 2 elements of group G

1. ## Proof of intersect between 2 elements of group G

Hi All

How do I prove the following:

Suppose that G is a group with elements a and b of orders p and q respectively, where p and q are distinct primes.

i) Show that <a> Intersect <b> = {e}, where e is identity element
ii)Suppose that for some intergers i and j, a^i = b^j. Show that p|i and q|j

Thanks again
The moolmiester returns

2. Is the first part of the question the same as the working in this one I found on the net?

27. Let G, G1, G2 be groups. Prove that if G is isomorphic to G1 × G2, then there are subgroups H and K in G such that
(i) H K = { e },
(ii) HK = G, and
(iii) hk=kh for all h in H and k in K. Solution: Let µ : G1 × G2 -> G be an isomorphism. Exercise 3.3.9 in the text shows that in G1 × G2 the subgroups
H* = { (x1,x2) | x2 = e } and K* = { (x1,x2) | x1 = e } have the properties we are looking for.
Let H = µ (H*) and K = µ (K*) be the images in G of H* and K*, respectively. We know (by Exercise 3.4.15) that H and K are subgroups of G, so we only need to show that
H K = { e }, HK = G, and hk=kh for all h in H and k in K.
Let y be in G, with y = µ (x), for x in G1 × G2. If y is in H K, then y is in H, and so x is in H*. Since y is in K as well, we must also have x in K*, so x is in H* K*, and therefore x = (e1,e2), where e1 and e2 are the respective identity elements in G1 and G2. Thus y = µ ((e1,e2)) = e, showing that H K = { e }. Since y is any element of G, and we can write x = h* k* for some h* in H* and some k* in K*, it follows that y = µ (h* k*) = µ (h*) µ (k*), and thus G = HK. It is clear that µ preserves the fact that elements of H* and K* commute. We conclude that H and K satisfy the desired conditions.

3. Originally Posted by moolimanj
i) Show that <a> Intersect <b> = {e}, where e is identity element
ii)Suppose that for some intergers i and j, a^i = b^j. Show that p|i and q|j
For the first question say $x\in \left< a\right> \cap \left< b\right>$ then by Lagrange's theorem $\mbox{ord}(x)|\mbox{ord}(a)|\mbox{ord}(a)$ and $\mbox{ord}(x)|\mbox{ord}(b)$ but $\mbox{ord}(a) = p$ and $\mbox{ord}(b)=q$ so $\mbox{ord}(x)$ divides both $p$ and $q$ thus $\mbox{ord}(x) = 1\implies x=e$ because $\gcd(p,q)=1$.

4. Thanks perfecthacker,

Can you help with the second part as well?

Cheers