Properties of Order of a Group Having real trouble getting started on this question, even though it doesn't seem hard:
Let $g,h \in G$ where $G$ is an Abelian group.
Then assume that $ord(g), ord(h)$ are finite with $hcf(ord(g),ord(h)) = 1$.
Prove $ord(g+h) = ord(g)*ord(h)$
I have tried showing that $ord(g+h)$ is the least positive integer $x$ such that $(g+h)x = 0$ and that this implies that $gx + hx = 0$.
Some help getting started would be great.
Thanks
 A: Because I'm on a phone, let $a$ be the order of $g$, and $b$ the order of $h$.
It suffices to show that $(g+h)$ has order $lcm(a,b)$. Look for powers that send $g+h$ to $0$. Because $g$ and $h$ commute, $(g + h)n = gn + hn$. So any common multiple of $a$ and $b$ works.
To show that only common multiples work, use the division algorithm. Say $gn + hn = 0$. We can divide $n$ by $a$ to get $n = aq + r, 0 \le r < a$. So $gn = g(aq) + gr = gr$. That part must be $0$, but $r < a$, so $r = 0$, and therefore $a$ divides $n$. Similarly for $b$.
So the $n$ that annihilate $g + h$ are exactly the common multiples of $a$ and $b$. The order is the least such $n$, so it's the lcm. Since $a$ and $b$ are relatively prime, their lcm is their product.
EDIT: Why does $gn + hn = 0$ imply $gn = 0$ and $hn = 0$? Raise both sides to $b$ (do you still say that with additive notation?): $g(bn) + h(bn) = 0$. But $h(bn) = 0$, so $g(bn) = 0$, and $a \mid bn$. But $a$ and $b$ are relatively prime, so $a \mid n$ and $gn = 0$.
A: Okay, so if $G$ is abelian, it is clear that $(g+h)^k=g^k+h^k$ (you can prove this inductively).  Consequently, $$(g+h)^{\text{ord}(g)\cdot\text{ord}(h)}=g^{\text{ord}(g)\cdot\text{ord}(h)}+h^{\text{ord}(g)\cdot\text{ord}(h)}=(g^{\text{ord}\,g})^{\text{ord}\, h}+(h^{\text{ord}\,h})^{\text{ord}\, g}=e+e=e$$
Thus $\text{ord}(g+h)|\text{ord}(g)\cdot\text{ord}(h)$.  Now, assume $(g+h)^k=e$ for some $k$.  Then $g^k=h^{-k}=(h^{-1})^k$.  $$\text{ord}(g^k)=\frac{\text{ord}\,{g}}{\gcd(k,\text{ord}\,g)}\;\;\;\text{ord}(h^{-k})=\text{ord}(h^k)=\frac{\text{ord}\,h}{\gcd(k,\text{ord}\,h)}$$
So $\text{ord}(g)\gcd(k,\text{ord}(h))=\text{ord}(h)\gcd(k,\text{ord}(g))$ and $$\text{ord}(g)|\text{ord}(h)\gcd(k,\text{ord}(g))$$
Since $\gcd(\text{ord}\,g,\text{ord}\,h)=1$, $\text{ord}(g)|\gcd(k,\text{ord}(g))$ and therefore $\text{ord}(g)|k$.  Similarly, $\text{ord}(h)|k$ and, since the orders are relatively prime, $\text{ord}(g)\cdot\text{ord}(h)|k$.  But $\text{ord}(g+h)|\text{ord}(g)\cdot\text{ord}(h)$, so $\text{ord}(g)\cdot\text{ord}(h)=\text{ord}(g+h)$.
QED
