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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 $H\neq \emptyset$?

Marios
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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). – George Law Jan 28 '17 at 00:19
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    A subgroup has to contain the unit element. – Bernard Jan 28 '17 at 00:19
  • 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 $H \neq \emptyset$... – Marios Jan 28 '17 at 00:23
  • @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. – Arthur Jan 28 '17 at 00:25
  • @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 $H \neq \emptyset$...???? – Marios Jan 28 '17 at 00:30
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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 $H=\emptyset$ – user190080 Jan 28 '17 at 00:32
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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? – Marios Jan 28 '17 at 00:36
  • mmmm interessant!! – Marios Jan 28 '17 at 00:44
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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 – user190080 Jan 28 '17 at 00:48
  • also habe ich richtig darüber gedacht... danke @user190080!! – Marios Jan 28 '17 at 00:50

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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 $x+(−x)$.

John B
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