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The standard definition of a Subgroup $H$ of a Group $(G,+)$ is as follows:

$H$ is Subgroup of $(G,+)$ if $\begin{cases} G \supseteq H\neq \emptyset \\ \forall x,y \in H:(x+y) \in H \\ \forall x \in H:(-x) \in H \end{cases}$

Why $H\neq \emptyset$?

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    Because a subgroup is a group in its own right, and groups are by definition nonempty (it must contain an identity element). – 2017-01-28
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    A subgroup has to contain the unit element. – 2017-01-28
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    but, $H$ is Subgroup than $\forall x \in H: (-x) \in H$ and $\forall x,y \in H: (x+y) \in H$ also $H \ni (x+(-x))=0 $ and $H \neq \emptyset$... – 2017-01-28
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    @Marios Actually, we have $\exists e\in H(\forall x\in H:ex = xe = x)$, rather than $H\neq\emptyset$. But in the end, with all the other axioms, either one implies the other. – 2017-01-28
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    @Arthur, also $H$ is Subgroup of $(G,+)$ if $\begin{cases} G \supseteq H\\ \forall x,y \in H:(x+y) \in H \\ \forall x \in H:(-x) \in H \end{cases}$ bacause I can prove $H \neq \emptyset$...???? – 2017-01-28
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    @Marios no, you can't prove this, you have to state this: the statement $\forall x,y \in H:(x+y) \in H $ is also true in the case of $H=\emptyset$ – 2017-01-28
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    @user190080, mmmmm if the statement $\forall x,y \in H:(x+y) \in H$ is not true in the case of $H=\emptyset$ than we have an absurd, namely $\exists x,y : x,y \in H=\emptyset \wedge (x+y) \notin H$... Is it correct? – 2017-01-28
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    mmmm interessant!! – 2017-01-28
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    @Marios maybe you find this interesting, especially the answer by Martin http://math.stackexchange.com/questions/50873/assumption-about-elements-of-the-empty-set – 2017-01-28
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    also habe ich richtig darüber gedacht... danke @user190080!! – 2017-01-28

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Actually, the assumption $H\ne\emptyset$ is made in the theory so that one can show that the subgroup of a group is also a group. But we neither assume from the beginning that it is a group nor for that matter that it has an identity. In other words, it is a consequence of the assumption $H\ne\emptyset$ that $H$ is a group and you can indeed use the argument given, but only if the set $H$ is nonempty since otherwise there is nothing to sum in $x+(−x)$.