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$\begingroup$

I have a problem that I am really confused on. It is the following:

Let $H$ be a subgroup of $(\mathbb{Z},+)$ generated by the integers $a_{1},a_{2},\ldots,a_{n}$. Show that $H=\{k_{1}a_{1}+k_{2}a_{2}+\cdots+k_{n}a_{n}\:|\:k_{1},k_{2},\ldots,k_{n}\in\mathbb{Z}\}$.

I'm just confused as to what needs to be proven. Isn't this simply the definition of a generated subgroup? I've already seen a proof that any subgroup of $(\mathbb{Z},+)$ is of the form $n\mathbb{Z}$ where $n\in\mathbb{Z}$. Do I need to use this fact anywhere for this proof?

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    Yes: this is simply the definition, you have nothing to prove. – 2017-02-02
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    ... Unless your definition of the subgroup generated by a set is the intersection of all subgroups containing the elements ("the smallest subgroup containing the generators"). This equivalence is a pretty popular exercise. – 2017-02-02
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    @rschwieb How would one actually go about proving this? – 2017-02-02
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    1. Show that $H$ is a subgroup. 2. Show that every subgroup containing $a_1, \ldots, a_n$ contains $H$. – 2017-02-02
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    The $n$ you mention is the $\operatorname{lcm}(a_1,\ldots)$. For example the subgroup generatied by $12,15$ and $20$ is $60\Bbb Z$. – 2017-02-03

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