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What is the order of $\mathbb{Z}_2\times\mathbb{Z}$?

For $\mathbb{Z}$ every non-zero element has infinite order, then $\mathbb{Z}$ has infinite order. Otherwise $\mathbb{Z}_2$ has order 2. Then what is the order of $\mathbb{Z}_2\times\mathbb{Z}$? Is infinite order?

Why $\mathbb{Z}$ has infinite order?

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    The order of a group is its cardinality – 2017-01-30
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    Beware: usually $\mathbb Z_2$ stands for $2$-adic integers. – 2017-01-30
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    @Santiago more often, I see $\Bbb Z_2 = \Bbb Z/2\Bbb Z$ – 2017-01-30
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    @Santiago Actually, I think that many mathematicians would assume that $\mathbb{Z}_2$ denotes the finite group of order $2$. – 2017-01-30
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    Right, the order of a group has nothing to do with the order of its elements. There is a relationship between those two for finite groups, but not visa versa. There are infinite groups whose non-identity elements all have order $2$, for example. Yet the group has infinite order. – 2017-01-30

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The order is the cardinality of the group, so the order of $\mathbb{Z}_2$ is 2 and the order of $\mathbb{Z}$ is infinite. Therefore the order of the cartesian product is also infinite

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    More precisely countable infinite. – 2017-02-01