Subgroups of finite and infinite groups Would somebody check the correctness of my proofs?
Theorem. Let $H$ be a non-empty subset of a finite group $G$. Then $H$ is a subgroup of $G$ iff $H$ is closed under the group operation.
Proof. If $H$ is a subgroup then it is closed. Suppsoe $H$ is closed.


*

*Associativity follows from $G$

*Since $H$ is finite, there exist $x,y\in  \mathcal N$ with $x\gt y$ such that $g^x=g^y$ where $g\in H$. So $g^{x-y}=e$ (the identity). $e=g^{x-y}\in H$ because it's closed.

*$gg^{x-y-1}=e$ so $g^{-1}=g^{x-y-1}\in H$ again because it's closed.


Therorem. Let $H$ be a non-empty subset of a group $G$. Then $H$ is a subgroup of $G$ iff $gh^{-1}\in H, \forall g,h\in H$
Proof. If $H$ is a sugroup then $gh^{-1}\in H$ by closure and inverses. Suppose that $gh^{-1}\in H$ for all $g,h\in H$.


*

*Associativity follows from $G$

*$gg^{-1}=e\in H$

*$eg^{-1}=g^{-1}\in H$

*$h^{-1}\in H$ for all $h\in H$, so $g(h^{-1})^{-1}=gh\in H$

 A: Some comments:


*

*I prefer "Associativity is inherited from G", although this is a personal opinion.

*In both proofs, you haven't used the non-emptiness of $H$.  "Since $H$ is non-empty, there exists $g \in H$."

*I'm uncomfortable with the possibility that $x-y-1=0$ in $gg^{x-y-1}$, without defining what $g^0$ is.

*You also seem to be getting a bit informal in the write-up of the second proof; what is $g$?, what is $h$?, etc.  Why not explain what you've shown at each step? "Thus, the identity element is in $H$." and "Thus, $g \in H$ implies $g^{-1} \in H$."

*I generally think that $\forall$ and $\exists$ should be used only in formal logic, or as a blackboard shortcut.

*Finally, to prove Theorem 2, having proved Theorem 1, you only need closure.
A: Your proof seems good; but the check about associativity is not necessary.
Here's how I like to prove this proposition.

What you have to prove is that $1\in H$ and that, for any $x\in H$, then $x^{-1}\in H$, under the hypotheses that $H$ is not empty and closed under the operation in $G$.
Since $H$ is not empty, there is an element $y\in H$. Define the map
$$
f\colon H\to H,\qquad f(x)=xy.
$$
This is well defined, by the closure of $H$ under the operation. Now $f$ is an injective map: indeed, if $x_1,x_2\in H$ and $f(x_1)=f(x_2)$, then $x_1y=x_2y$, whence $x_1yy^{-1}=x_2yy^{-1}$ (this can be done in $G$, which is a group).
Since $H$ is finite, we conclude that $f$ is surjective. In particular there exists $e\in H$ such that $f(e)=y$, that is $ey=y$, which means $e=1$.
We have thus proved that $1\in H$; so there is $z\in H$ with $f(z)=1$, which means $zy=1$, that is, $z=y^{-1}\in H$.
Since $y$ is an arbitrary element of $H$, we are done.

Note that we don't need $G$ to be finite.
