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In this SE answer about showing that $\mathbb Z _6 \times \mathbb Z _2 $ doesn't have any element of order $12$, we have $(x,y)\in \mathbb Z _6 \times \mathbb Z _2$. It says that $o(x) \mid 6$ and $o(y) \mid 2$. It hence says that $lcm(o(x),o(y))$ is at most $lcm(2,6) = 6$, which I don't get. Here $o(x)$ denotes the order of $x$.

So far, here is my attempt at finding out why $o(x) \mid 6 \land o(y) \mid 2 \implies lcm(o(x),o(y)) = 6 \text{ at most}$:

I tried to consider the $gcd$ of $o(x)$ and $o(y)$: $(o(x),o(y))$ $$ (o(x),o(y))\mid o(x),o(y) $$ $$ \implies (o(x),o(y)) \mid 6,2 $$ $$\implies 2,6 \mid [o(x),o(y)] $$ the last line being obtained by $(o(x),o(y))[o(x),o(y)] = o(x)o(y)$ where $[o(x),o(y)]$ denotes the $lcm$ of $o(x)$ and $o(y)$.

if anyone could give me a hint at how to do this, I would greatly appreciate. Thank you in advance.

Bill Dubuque
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niobium
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1 Answers1

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If $b\mid 2$, then we also have $b\mid 6$. And since $a\mid 6$ and $b\mid 6$, then we immediately have that $6$ is a common multiple of $a$ and $b$. Then it's obvious that $12$ can't be the least common multiple.

Arthur
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