Inverses in subspace? Why do we not have to prove that additive inverses exist in order for a subset of a vector space to be a subspace whereas in the case of groups we do? Does the existence of inverses come about from closure under scalar multiplication?
 A: By the most logical definition, a subspace of a vector space $V$ is a subset $U$ of that is a vector space under the same addition and scalar multiplication (restricted, of course). So in principle you have to prove all the axioms of a vector space. But by the very fact that $V$ is already known to be a space, a lot can be left out, for example, associativity of addition need not be checkd in $U$ because iot holds in $V$. Similarly, we need not check that $-u \in U$ for all $u\in U$, provided we already know that $cu\in U$ for all $u\in U$, $c\in F$. This is so because we are allowed to let $c=-1$ and already know (in $V$) that $(-1)\cdot u=-u$. So what we have is a simplified test whether $U$ is a subset of $V$, or a subspace criterion: We need only check that $U$ is not empty, that it is closed unde addition, and closed under scalar multiplication. Essentially, we have just shown most of the claim that any subset with these properties is a subspace.
There is a similar situation with groups: A subgroup is a subset of a group that happens to be a group under the given (restricted) group operation. Again, we do not have to check all properties, e.g., associativity is - again - for free. The lack of scaler multiplication seems to require that we always have to show the closure under taking inverses, but in fact this is not always the case. There is a subgroup criterion for the special case of finite groups which only requires one to check that the presumed subgroup is non-empty and closed under the group operation!
