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What is meant by "set of all integral multiples of 6"

I'm working on group theory. And trying to disprove or prove this is a group. But couldn't understand what is meant here. Integral multiples?

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    $6\Bbb Z$ is just the set of all integers times $6$. So $6(0)=0$ is in there, $6(1)=6$ is in there, $6(-1)=-6$ is in there, and so on. – 2017-02-26
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    $...-36, -30, -24, -18 \cdots 0 \cdots 18, 24, 30, 36$ Get the pattern? – 2017-02-26
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    To be formal, it means $\{6x : x \in {\mathbb Z}\}.$ – 2017-02-26
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    The given set with the usual addition is a group. $0$ is the neutral element, the inverse of $x$ is just $-x$ and the law of associativity is satisfied because we are adding integers. And finally, if we add two elements, we are landing again in the set. – 2017-02-26

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Perhaps the confusion comes from the word 'integral'. Here it does not mean $\int$ but means multiply by an integer. So the set you are working with is all integers $\{ \ldots, -2, -1, 0, 1, 2, \ldots \}$ multiplied by 6 and of course that means multiply every element of that set by 6.