# Is such set a group?

Question: If a set $G$ is equipped with an associative binary operation $\ast$, and assume $G$ has identity element $e$ and for each $g \in G$ there exists its inverse element $g^{-1}$, is $G$ a group?

Here is the original question and solution:

I have done pretty much the same job as above, but like the solution, there is no check on closure of the operation $\ast$. Actually I don't think we can do that because for example $\{-1, 0, 1\}$ as a subset of $\Bbb Z$ with the same operation of $+$ in $(\Bbb Z, +)$ is not closed under it.

I am so frustrated and could someone please explain?

• Your question in yellow does not match the original question. – Matt Samuel May 13 '15 at 4:22
• having a binary operation on a set means the set is closed under the operation. That is the meaning of being a binary operation on a set. – Ittay Weiss May 13 '15 at 4:23
• Indeed, the question in yellow has an obvious affirmative answer because this is the definition of a group. Closure is implicit; otherwise there is no such binary operation. – Matt Samuel May 13 '15 at 4:24
• @Ittay Weiss Oh it seems I misunderstood the concept. Thank you very much. – MonkeyKing May 13 '15 at 4:29
• I just want to make a small note about the way you phrased a portion of your question (I know I'm nitpicking). You said there exists an inverse for all $g$. Because of the way you ordered the quantifiers ($\exists g^{-1} \forall g$), you are saying there is a single element that acts as the inverse for every element in the group. Actually, different elements can have different inverses, so the correct way to phrase the statement is: for each element, there exists an inverse (i.e., $\forall g, \exists g^{-1}$). – layman May 13 '15 at 4:32

By definition a binary operation is a map from $G\times G \to G$ so it is closed. If in addition to that set also has an identity for that operation, and every element has an inverse, then you've got a group my friend.