What do I need to show that a subset of a group is a subgroup? I am working through Fraleigh's text in abstract algebra.
What things must I show to prove that a subset H of a group G is a subgroup under the group operation? 
I thought I had to show:
1) The identity from the group is the identity for the subgroup and is in the subgroup
2) The group is closed under inversion (group operation with inverse element) and that the inverse for the group is the inverse for the subgroup
3) The group is closed under the group product
If I am wrong, please correct me with the proper approach.
Also, what is the easiest way to show that the identity / inverse from the group are the same as in the subgroup? This seems trivial enough as to be difficult.
 A: That's correct, those are the things you need. But there are some shortcuts - regarding the first one, note that no two elements of a group behave like the identity. What I mean is that if $h \in G$ isn't an identity element and $xh = h$, then $x$ is the identity element. That means that in order for $H$ to have an identity element, it has to include the original identity element of $G$ - so all you need to do is check whether that's in $H$.
Likewise, nothing has two inverses - so if $h$ has an inverse in $H$, that's automatically the same inverse as in $G$. So all you really need to do is this:
1) Check whether $e \in H$, where $e$ is the identity element of $G$.
2) Check whether, for each $h \in H$, $h^{-1}$ is also in $H$, where $h^{-1}$ is the inverse of $h$ in $G$.
3) Check whether, for each $h_1,h_2 \in H$, $h_1h_2$ is also in $H$.
If the answer to all three is "yes", then $H$ is a subgroup.
A: Given a group $G$, with its binary ("product") and unary ("inverse") operations:
$$
\begin{equation}
\begin{split}
\circ&\colon G^2\to G\\
\operatorname{inv}&\colon G\to G
\end{split}
\end{equation}\tag{1}
$$
you can consider the restrictions of them on $H^2$ and $H$ respectively, where $H\subset G$:
$$
\begin{equation}
\begin{split}
\circ_{H^2}&\colon H^2\to G&;~(h_1,h_2)&\mapsto h_1\circ h_2\\
\operatorname{inv}_H&\colon H\to G&;~h&\mapsto\operatorname{inv}(h)
\end{split}
\end{equation}\tag{2}
$$
When you ask whether $H$ is a subgroup of $G$, you are asking whether $H$ is a group with the group structure induced by that of $G$, that is, with binary and unary operations pointwise concident with those of $G$ on $H^2$ and $H$ respectively, that is, with binary and unary operations pointwise coincident with the restrictions $(2)$ of the corresponding operations of $G$. In the end you want to know whether there exist corestrictions to $H$ of $(2)$'s, or directly birestrictions of $(1)$'s to the pairs $(H^2,H)$ and $(H,H) $ respectively:
$$
\begin{equation}
\begin{split}
\circ_{H^2}^{H}&\colon H^2\to H&;~(h_1,h_2)&\mapsto h_1\circ h_2\\
\operatorname{inv}_H^{H}&\colon H\to H&;~h&\mapsto\operatorname{inv}(h)
\end{split}
\end{equation}\tag{3}
$$
These existence conditions are called closure of $H$ under product and inversion.
There remain to be proved associativity, the existence of identity and its equality to the identity of $G$. But the associativity is a pointwise property of $\circ$ that a fortiori holds when $\circ$ is restricted or birestricted. Moreover if the identity of $H$ exists it must be equal to that of $G$ because the identity of $G$ is defined by a pointwise property of $\operatorname{inv}$ that a fortiori holds when $\operatorname{inv}$ is restricted or birestricted.
But you have no way to prove for the general case that the identity of $H$ exists unless you add to your hypotheses that $H$ must be non-empty.
This last was the only important thing missing in your reasoning.
