Lemma to FToFinite Abelian Groups, clarification 
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I am trying to understand this Lemma 8.4's proof. I understand what is happening until just after the line that says
$$a=1a=(mu+nv)a=mua+nva.$$
At this point I do see why $mua \in G(p_1)$ since $n(mua)=u(mna)=u(0)=0$ so $\mathrm{ord}(mua) \mid n=p_1^{r_1}$ so $mua \in G(p_1)$. What I am wondering now is why even mention this? It looks like the next step that is taken is saying $n(mua)=u(mna)=u(0)=0$, but this doesn't seem to depend on $mua \in G(p_1)$, it seems more to just depend on $\mathrm{ord}(a)=mn$. I see too why $m(nva)=0$ as $m(nva)=v(mna)=v(0)=0$ so indeed, $\mathrm{ord}(nva) \mid m = p_2^{r_2}p_3^{r_3}\cdots p_k^{r_k}$.
My trouble now at this point in the proof is concluding then that $\mathrm{ord}(nva)$ has $(k-1)$ distinct prime divisors. I see that $\mathrm{ord}(nva)$ divides a number that consists of $(k-1)$ distinct prime factors but I guess I don't see why it follows that $\mathrm{ord}(nva)$ has $(k-1)$ distinct prime divisors. Regardless, is it just like we know for sure that there are at most $(k-1)$ prime divisors of $\mathrm{ord}(nva)$ by transitivity of $``divides''$ and then if it in fact has less we just say some of those $a_i =0$?
Some of my thinking may not be in the right direction, could someone help to clarify this proof somewhat or is there another proof I can look at?
Thanks
 A: The answer to your first question, "Why even mention this?" [that $mua\in G(p_1)$] is that this fact is used in the last line of the proof.  There, $a_1$ is defined to be $mua$ and it is asserted that $a_i\in G(p_i)$ (for all $i$, including $i=1$).
The simplest answer to your second question, why the order of $nva$ actually has $k-1$ distinct prime divisors and not fewer, is that it would do no harm to the proof if $nva$ had fewer than $k-1$ prime divisors.  The reason is that the induction hypothesis is that the result holds "for all elements whose order is divisible by at most $k-1$ primes."  I suspect that, where the author wrote that $nva$ has "only" $k-1$ distinct prime divisors, the intended meaning was that it has at most that many.  
Nevertheless, it is in fact true that $nva$ has $k-1$ distinct prime divisors.  If it had $k-2$ or fewer, then a product of sufficiently high powers of those dvisors would annihilate $nva$.  Since a sufficiently high power of $p_1$ annihilates $mua$, we would have that the product of these two annihilating multipliers, which has only $k-1$ or fewer prime factors, would annihilate $mua+nva=a$, contrary to the assumption that the order of $a$ has $k$ distinct prime divisors.
A: First of all, concerning your first remark on the proof, about $n(mua) =u(mna) = 0$ not depending on $mua$ dividing $p_1^n$ for some $n \geq 0$, you are right. The wording was a little bad. What the author wanted, was to show that $mua$ was in $G(p_1)$, not to use it as an unproven result to prove something else. 
He/she should have said: $mua \in G(p)$, because $a$ has order $mn$ and therefore we must have $nmua = 0$.
As for the second problem:
Notice that $0$ is in every $G(p_i)$. The author says the order of $m$ has $k-1$ distinct primes, not that the order of $nva$ has that. Else, see what Andreas wrote.
By the way, what book is that?
