Proof of LCM property.

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I have to prove that $$LCM(a,c)|LCM(ak,cm)$$ For some $k,m \in \Bbb Z$, I've been going around this some time, but cannot come to anything, any help or hint would be great! thanks!

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Yes: $a\mid ak\implies a\mid\operatorname{lcm}(ak,cm)$. For the same reason, $c\mid\operatorname{lcm}(ak,cm)$. Therefore, $\operatorname{lcm}(a,c)\mid\operatorname{lcm}(ak.cm)$.