# If $p_i$ are distinct primes, show that an abelian group of order $p_1p_2…p_s$ must be cyclic.

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 $$|| = p_ip_j$$.

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

The proof is correct, but you should justify why $$|\langle p_1'p_2'\rangle|=p_1p_2$$.
• 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 $|<p'_ip'_j>|=p_ip_j$? – Mitchell Faas Oct 2 '19 at 11:42
• 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 $n=k*lcm(p_i, p_j)$. – Mitchell Faas Oct 2 '19 at 11:47
• 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. – QuantumSpace Oct 2 '19 at 11:49