Suppose the set $G$ is an additive group of integers $(G,+)$.
For a subset $H$ of the set $G$ to be a subgroup,
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the subset $H$ must contain the identity element
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the subset $H$ must be closed under the group binary operation
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the subset $H$ must contain inverses of each of its elements
1) I generally have problem showing the identity element exists even though it is purported to be the easiest way to prove.
Suppose I have shown that condition 2 and 3 holds, would adding the element of the subset $H$ to its inverse suffice to show that the identity element exists?
2) Would I have to show all of these condition holds or does showing just one condition suffice to show that the subset $H$ of a group $G$ is a subgroup.
Best Answer
If $H$ is closed under multiplication and contains the inverse of each of its elements, then it contains also the identity as long as $H$ is non-empty.
Indeed $$ H\neq\emptyset\Rightarrow \exists h\in H\Rightarrow \{h,h^{-1}\}\subset H\Rightarrow 1=hh^{-1}\in H. $$