Question on subgroups "inheriting" the group operation Suppose $G$ is a group under addition. Must it be the case that the subgroup $H$ inherits the same product operation as $G$? For example, if $G = \mathbb{R}$ and $H$ is a proper subset of $\mathbb{R}$ with a group structure under multiplication, is $H$ still a subgroup of $G$? 
This topic doesn't seem to be covered in any of the textbooks I am working in, perhaps because it is considered obvious. In this particular case, the identity and inverse elements would differ, but in theory there could be a case where the operation differs but the identity and inverses are preserved. (Or is there?) 
I suppose I am curious on whether the product operation is preserved by definition or by necessity (i.e., we can't find an example where it isn't) or whether this isn't even a requirement to begin with.
 A: A subgroup must inherit the same group operation as the supergroup, as a matter of definition. For example, one could easily define modulo $4$ arithmetic on the subset $\{0, 1, 2, 3\}$ of $\Bbb{R}$ (under addition), but it's not a subgroup, because $3$ and $1$ do not add to $0$ in $(\Bbb{R}, +)$, but they do in modulo $4$ arithmetic.
One could do something similar with $\{1, 2, 3, 4\}$, or $\{8, -1, \pi, 2.1\}$. Note that the group structure of $(\Bbb{R}, +)$ has nothing to do with it. Really, the only structure necessary to form this relaxed type of "subgroup" is for the set to have cardinality less than or equal to $\Bbb{R}$. That is, the formation of these "subgroups" is less about how they interact via $+$ and more about the number of elements in $\Bbb{R}$.
A: A subgroup $H \subset G$ has to have the same group operation as $G$. You could define on the set $H$ a new operation which turns $H$ into a group (see the other answer) - but it wouldn't be a subgroup of $G$ anymore. 
The idea is the following: If you have two sets $A, B$ with $A \subset B$  you think of the set $A$ being contained in $B$. But a group $G$ is not only a set , it's a set $G$ together with a group operation $+$, so it's $(G,+)$. As before we want a subgroup $H$ of $G$ to be contained in $G$. But because we know think about groups and not sets anymore it's not enough for $H$ to be simply contained as a set in $G$. We want $H$ to be contained as a group in $G$ and that means that it has to have the same group operation (and is itself a group). 
You will find sub-somethings often. Subspaces for vector spaces, subrings for rings and so on. The idea is always the same. A sub-something should not only be a subset but a sub-something. So a subspace $U$ of a vector space $V$ should not only be a subset $U \subset V$ but it should be a subspace, meaning it should be itself a vector space and the vector space operations stem from $V$. So it's contained in $V$ as a vector space and not as a set.
