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So I, more or less, need help understanding this problem.

The first thing I have tried is to pick a specific group: $\mathbf{Z}_5$, under multiplication.

To demonstrate, say we pick element $2$ and $|2| = 4$, and then the only other possible orders are $\{1,4\}$.

Indeed, $4 = LCM(1,4)$.

Not really quite sure, how to generalize this, yet. Any hints? What shall I think about? Definitely prime factorization of each element, hmm... I think!

user1729
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novicemathsdancer
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    Do you know the fundamental theorem of finite abelian groups? – Jo Mo Jun 27 '18 at 06:40
  • That's too simple a case to give much insight into the problem. My hint is to consider elements of prime-power order. – Angina Seng Jun 27 '18 at 06:40
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  • If it's under multiplication, it ought to be written $\Bbb Z_5^\star$ as the set of invertible elements of $\Bbb Z_5$. – Arnaud Mortier Jun 27 '18 at 06:47
  • Yes, the Fundamental Theorem of Finite Abelian Groups guarantees some elements of prime power order. You can certainly construct an element with the required $\text{lcm}$ order. –  Jun 27 '18 at 07:04
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    Thanks @GerryMyerson, not possibly, it is the exact problem! However, I wanted to see if I can get there on my own without actually seeing a proof, hence, I asked for hints. – novicemathsdancer Jun 27 '18 at 07:13
  • OK, then, hint: prove that if the orders of $a$ and $b$ are relatively prime, then the order of $ab$ is the product of the order of $a$ and the order of $b$. – Gerry Myerson Jun 27 '18 at 07:16
  • @GerryMyerson thanks for that, I can use Bézout's identity to show that product = lcm / gcd, since gcd = 1, here, it follows that product = lcm. Now, I have to show that every possible order in a finite abelian group are all relatively prime...this might have something to do with that totient function, but how do I know that that's all there is for finite abelian groups? – novicemathsdancer Jun 27 '18 at 07:45
  • No, the possible orders are not all relatively prime. But suppose $a$ has order 12, and $b$ has order 18 – can you figure out a way to get elements of orders 4 and 9? – Gerry Myerson Jun 27 '18 at 08:58
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    @GerryMyerson, yep, just take $a^3$ and $b^2$. – novicemathsdancer Jun 27 '18 at 16:06
  • @GerryMyerson I think...by definition of order I can use Fermat's Little Theorem to say that the orders of the elements in the group would share the same prime power? or am I thinking of a too specific case? I have $\mathbf{Z}_{19}$ in mind for this example. – novicemathsdancer Jun 27 '18 at 16:30
  • @GerryMyerson also, another helpful fact: if the order of element $a$ is $d$ then $d|n$ in $a^n = 1$ so I can work with this backwards? – novicemathsdancer Jun 27 '18 at 17:16
  • Fermat is a theorem about integers modulo a prime. Not every finite abelian group is of that type, so Fermat is irrelevant here. And, yes, if $a^n=1$ then $n$ is a multiple of the order of $a$. – Gerry Myerson Jun 27 '18 at 22:20

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