1. Group Theory Proof

Can somebody help me with this proof please?

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

Suppose also that for some integers i and j, that a^i = b^j. Show that p|i and q|j.

2. Hi
Originally Posted by Cairo
Suppose that G is a group with elements a and b of orders p and q respectively...
This means that $a^p=b^q=e$ where $e$ is the identity element of the group.
... where p and q are distinct primes.

Suppose also that for some integers i and j, that a^i = b^j. Show that p|i and q|j.
You may try using $(a^i)^p=(b^j)^p$ and $(a^i)^q=(b^i)^q$.

3. could be having a dumb moment....but I'm still not following.

4. As $(a^i)^p=a^{ip}=(a^p)^i=e^i=e$ and $a^{ip}=b^{jp}$, we get that $e=b^{jp}$ which requires that there exists $n_1\in\mathbb{N}$ such that $jp = n_1 q$. What can be deduced from this ?

5. that p|i ?

6. Bad luck, it's the other one... (it can be shown acknowledged that if $u|vw$ then $u|v$ or $u|w$)