Proving Theorem on Sub-Group Consider the below theorem on Sub-Groups

Theorem : Let $(G, \circ)$ be a group and $H$ be a non-empty finite subset of $G$. Then $(H, \circ)$ is a subgroup of $(G, \circ)$ if and only if $a \in H, b \in H \implies a \circ b \in H$

Proving Forward : If $(H, \circ)$ is a subgroup of $(G, \circ)$, then $a \in H, b \in H \implies a \circ b \in H$ using Closure Property.
Proving Backward : If $a \in H, b \in H \implies a \circ b \in H$, then $(H, \circ)$ is a subgroup of $(G, \circ)$.
In order to prove this we have to satisfy 4 properties of group i.e.


*

*Closure

*Associative

*Existence of Identity

*Existence of inverse


Property 1: Satisfied
Property 2: Satisfied. Since $H$ is a non-empty subset of $G$ and $\circ$   is associative on $G$, $\circ$ is assoicative on H.
Property 3: ?
Property 4: ?
How can I prove existence of identity and existence of inverse with just information of Closure Property.
 A: Since $H$ is non-empty, pick any old $a \in H$. Consider the map; $H \to H$ given by:
$h \mapsto ah$ (left-multiplication by $a$).
This map is injective (since it's injective on all of $G$), and since $H$ is finite, it's also surjective on $H$ (can you think of an example where this fails for some group $G$ and an infinite subgroup $H$?).
So there must be some $h_0 \in H$, with $ah_0 = a$ (it might be $a$ itself).
Similarly, the map $h \to ha$ is also surjective. This means for any $b \in H$, we have some $x \in H$ with $xa = b$.
Thus $bh_0 = (xa)h_0 = x(ah_0) = xa = b$, for any $b \in H$.
A similar argument shows that there is an $h_1 \in H$ with $h_1a = a$, and for any $c \in H$, another element $y \in H$ such that $ay = c$.
Hence $h_1c = h_1(ay) = (h_1a)y = ay = c$, for any $c \in H$.
So we have a left-identity, and a right-identity. Thus:
$h_1 = h_1h_0 = h_0$.
Thus $h_1 = h_0$ is an identity for $H$. Now it is easy to see that any element $H$ has both a left-inverse, and a right-inverse (just consider the pre-image of the element $e$ under the maps $x \to hx$ and $x \to xh$). So if we suppose $x$ is a left inverse for $h$, and $y$ a right-inverse:
$x = xe = x(hy) (xh)y = ey = y$
so two-sided inverses exist for any $h \in H$, and being unique in $G$, must be unique in $H$.
A: For any $a\in H$, $\;\{a^n\mid n\in\mathbf N\}\subset H $. Hence this set is finite, so there are exponent $i,j,\enspace i<j$, such that $a^i=a^j$. As we're in a group, the cancellation law gives $a^{i-j}=e\in H$, whence $a^{-1}=a^{i-j-1}\in H\;$ (note $i-j-1\ge 0$).
