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I've got a (possible) solution for this problem, but I'm not very satisfied with how I've written it down and would like some feedback.

If $G$ is such a group, then by Cauchy we know that there exist elements in $G$ of order $p_i,\;1\leq i\leq s$; denote these elements as $g_i$.

Because of Lagrange, we know that these elements must be generators for groups of order $p_i$ respectively.

Let $g_i, g_j$ distinct. Then because $G$ is abelian, $(g_ig_j)^{n} = g^n_i g^n_{j}$. So $(g_ig_j)^n=e$ if and only if $g_i^n=e$ and $g_j^n=e$. Therefore the co-primeness of their orders ($p_i$ and $p_j$ are both distinct primes) ensures that $|<g_ig_j>| = p_ip_j$.

Continuing this process, $|<g_1g_2...g_s>|=p_1p_2...p_s$, so $g_1g_2...g_s$ is a generator of the entire group.

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The proof is correct, but you should justify why $|\langle p_1'p_2'\rangle|=p_1p_2$.

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  • $\begingroup$ How would you best write this down? Since $p_i$ and $p_j$ are distinct primes, their least common multiple is $p_ip_j$, and therefore $|<p'_ip'_j>|=p_ip_j$? $\endgroup$ – Mitchell Faas Oct 2 '19 at 11:42
  • $\begingroup$ What has the least common multiple have to do with this? (What theorem are you using?) $\endgroup$ – QuantumSpace Oct 2 '19 at 11:43
  • $\begingroup$ I'm not really using any theorems for this step, just recognising that the order of $p'_i$ and $p'_j$ are $p_i$ and $p_j$, so if $(p'_ip'_j)^n=e$, then $p'^n_i=e$ and $p'^n_j = e$, so $n=k*lcm(p_i, p_j)$. $\endgroup$ – Mitchell Faas Oct 2 '19 at 11:47
  • $\begingroup$ That works. Just add this little explanation in the proof and you'll be done. I think the proof looks a little bit awkward because you call the elements $p_i'$. Rather call them $g_i$ or something like that. Makes it less confusing to read and it is immediately clear that the elements $g_i$ are group elements. $\endgroup$ – QuantumSpace Oct 2 '19 at 11:49
  • $\begingroup$ So like this? :) $\endgroup$ – Mitchell Faas Oct 2 '19 at 11:56

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