Suppose the set $G$ is an additive group of integers $(G,+)$. For a subset $H$ of the set $G$ to be a subgroup,

  1. the subset $H$ must contain the identity element

  2. the subset $H$ must be closed under the group binary operation

  3. 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.


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 $$ H\neq\emptyset\Rightarrow \exists h\in H\Rightarrow \{h,h^{-1}\}\subset H\Rightarrow 1=hh^{-1}\in H. $$

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  • $\begingroup$ What if H is closed under addition? $\endgroup$ – Mathematicing May 17 '15 at 13:42
  • $\begingroup$ Suppose G={g1,g2,....} and H={2g1,2g2,....} Then, the it seems the identity element can be shown to exists if 2g1+(-2g1)=0. Since 2g1 is in the subset H and we have shown that -2g1 is the inverse of 2g1, does it not suffice to say the identity element exists if 2g1+(-2g1)=0? $\endgroup$ – Mathematicing May 17 '15 at 13:46
  • $\begingroup$ @user: groups have only one operation. To call it multiplication and denote it accordingly is merely conventional. I'm afraid I don't understand your other question. $\endgroup$ – AdLibitum May 17 '15 at 13:53

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