Let $(R,+,\cdot)$ be a non-commutative ring with zero $0$ and identity $1$. $a,b\in (R,+)$ are two elements of finite order $m,n$ respectively; in notations, $ord(a)=m,ord(n)=n$. Then what is the order of the product $ab\in (R,+)$ ? In particular, if $m=n$, then is it ture that $ord(ab)=n?$
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$R$ is commutative. Assuming you mean multiplicative order.
If $a$ has order $n$ then $a^n = 1$ and therefore $a^k=1$ whenever $k$ is a multiple of $n$. Similar formulation for $b$ of order $m$. Then $$ (ab)^k = a^k b^k = 1 $$ provided $k$ is a multiple of both $n$ and $m$. So the order of $ab$ is at most $\mathrm{lcm}(n,m)$.
The order of $ab$ could be smaller than that. For example, if $a$ is the inverse of $b$, then the order of $ab$ is $1$.
Of course if you mean additive order, this is not the answer. Which is why you should fix your problem to say what order you mean.
I believe you meant additive order... did you try even a single example? Like, say $\mathbb Z/4\mathbb Z$?
$ord(2)=2$ and $ord(3)=4$, but $ord(2\cdot 2)=ord(0)$ and $ord(2\cdot 3)=ord(2)$.
In other words, the order of the product can be strange things compared to the order of the factors. I'm sure you can go about finding other nonintuitive examples just using $\mathbb Z/n\mathbb Z$.