Talk:Group Direct Product of Cyclic Groups

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Any objections to changing this to an "iff" statement? --Linus44 (talk) 13:28, 11 May 2013 (UTC)

Yes because the reverse does not necessarily hold, from what I understand. I think that a group of order 21 is not cyclic. I may be wrong, I need to go check. --prime mover (talk) 14:25, 11 May 2013 (UTC)
Sorru, I'm talking complete rubbish there. Let me go away and think what I'm saying. --prime mover (talk) 14:27, 11 May 2013 (UTC)
... What actual statement are you proposing? Where does the "iff" come in? --prime mover (talk) 14:31, 11 May 2013 (UTC)

As follows:

Theorem

Let $G$ and $H$ both be finite cyclic groups.

Let $g = |G|, h=|H|$ be their respective orders.

Then their group direct product $G \times H$ is cyclic if and only if $g$ and $h$ are coprime, i.e. $g \perp h$.

Proof

$\Leftarrow$: already done.

$\Rightarrow$: Let $(x,y)$ generate $G \times H$.

Then $x$ generates $G$ and $y$ generates $H$, so $|x| = g$, $|y| = h$.

By Order of Group Element in Group Direct Product, $gh = |(x,y)| = \operatorname{lcm}(|x|,|y|)$.

Therefore we have $gh = \operatorname{lcm}(g,h)$ and therefore $\gcd(g,h) = 1$ by Product of GCD and LCM.

$\blacksquare$

I propose the change in part because I couldn't figure out how to name the converse in relation to this result. --Linus44 (talk) 15:04, 11 May 2013 (UTC)

O yes of course. Makes sense. I was getting confused with non-cyclic groups and other irrelevant stuff. Go for it. --prime mover (talk) 16:48, 11 May 2013 (UTC)