Order of a $\alpha \beta^{n/q}$ given the order of $\alpha$ and $\beta$ I'm asked to prove the following easy result:
Let $G$ be a finite abelian group. Let $\alpha \in G$ of order $m$ and $\beta \in G$ of order $n$. Assume the $n\not\mid m$, and let $q=p^v$ for some prime $v$ the greatest power of $p$ s.t. $q\mid n$ but $q \not\mid m$. Consider the element $$\alpha\beta^{n/q}$$ and prove it has order the l.c.m. of $m$ and $q$.
It's easy to see that the order of $\alpha\beta^{n/q}$ (call it $k$) divides $l.c.m(m,q)$. But I'm getting confused in proving that it has to be the l.c.m. In fact we have $$ 1=\alpha^k\beta^{kn/q}$$ which means $$ \alpha^{-k}=\beta^{kn/q}$$
But then I don't know which manipulation should I do to get the result I want
can someone provide any hints?
 A: the previous version of this answer had an issue, hope that now is correct:
Call $\tilde{\beta}$ the element $\beta^{n/q}$. By construction it has order $q=p^v$. By hypothesis $m=p^rm'$, with $r<v$ and $(m',p)=1$.
By the equality $$ \alpha^{-k}=\tilde{\beta}^k$$ we have that $m\mid kq$ and $q \mid km$ in fact if we raise both side to the $m$-th power say, we would have $1=\tilde{\beta}^{mk}$ and therefore the order of $\tilde{\beta}$ has to divide $km$ as claimed. In particular, $$m'=\frac{m}{(m,q)}\mid k $$ and $$p^{v-r}=\frac{q}{(m,q)} \mid k$$
Notice that by construction $(m',p^{v-r})=1$, and that $l.c.m(m,p^v)=m'p^v$ therefore by the property of the lcm, the last two divisions imply that $$m'p^{v-r}\mid k \mid m'p^v$$  So $k=m'p^s$ for some $v-r\leq s \leq v$. Now we have two cases: 
$s<r$, which leads to $\alpha^{m'p^s}=\beta^{\frac{n}{p^{v-s}}m'}$ and by using the coprimality of $m'$ and $p$ we conclude that (checking the orders of the two elements $r=v$, an absurd.
$s\geq r$ which gives us $1=\beta^{\frac{n}{p^{v-s}}m'}$ and therefore $v=s$ and so we have proved the claim about $k$.
