Are there anything wrong in the following proofs

Question

Problem 1

Let $\left(S,\circ\right)$ be a semigroup. If for any $a$ and $b$ the equations $a\circ x=b$ and $y\circ a=b$ has a solution then show that $\left(S,\circ\right)$ is a group.

Proof

We use the following definition of a group.

A binary structure $\left(G,\circ\right)$ is said to be a group if,

<ol>
<li><p>$(a\circ b)\circ c=a \circ (b\circ c)\ \ \forall a,b,c\in G$</p></li>
<li><p>$\exists e \in G \mid e\circ x=x \ \ \forall x\in G$</p></li>
<li><p>$\exists z \in G \mid z\circ a=e \ \ \forall a\in G$</p></li>
</ol>

$ \exists x,y\in S \mid(a \circ x=b) \land (y \circ a=a)\implies y\circ (a \circ x)=y\circ b \implies (y \circ a)\circ x=y\circ b \implies a \circ x=y\circ b \implies b=y \circ b$

This shows that the same $y$ which satisfies the equation $y \circ a=a$ also satisfies $y \circ b=b$ for all $a, b \in S$. Hence (2.) is satisfied.

Now we call this element $y$ by $e$ and find the solution of $z \circ a=y$. Since by problem such a $z$ always exists, (3.) is also satisfied.

Hence $(S, \circ)$ is a group.

Problem 2

$H\leq G \iff a\circ b^{-1} \in H\ \ \forall a,b \in H$

Proof

Now the part $H\leq G \implies a\circ b^{-1} \in H\ \ \forall a,b \in H$ is trivial since $H \leq G$ implies that there is a unique solution of the equation $x \circ b=a\ \ \forall a,b \in H$ and hence the proof is skipped. Only the proof of the second part is discussed below.

To prove closure in $H$ first of all we prove the existence of identity.

$a\circ b^{-1} \in H \implies a\circ a^{-1} \in H \implies e\in H$

Notice that this is the same $e$ of $G$. Hence existence of identity is confirmed.

Then $a,e \in H \implies e \circ a^{-1} \left(=a^{-1}\right)\in H$ Thus existence of inverse is also confirmed. Thus $a,b \in H \implies a,b^{-1} \in H \implies a\circ b \in H$. Hence closure is also established.

The thing that remains to be shown is associativity. $a,b,c \in H\implies a,b,c \in G \implies (a\circ b) \circ c=a \circ (b \circ c)$

Hence the result is proved.

Are there anything that I have missed?

Answer 1

Your proof for problem 1 works, but when you write

$ \exists x,y\in S \mid(a \circ x=b) \land (y \circ a=a)\implies y\circ (a \circ x)=y\circ b \implies (y \circ a)\circ x=y\circ b \implies a \circ x=y\circ b \implies b=y \circ b$

you should quantify $a$ and $b$, e.g. "$\forall a,b \in S, \exists x,y\in S ~\mid$ ..."

In problem 2, the reverse implication is false if $H$ is the empty set, so there is no proof for the problem as stated. This shows up in your proof when you write

$\dots \implies a\circ a^{-1} \in H \implies \dots$

which assumes $\exists a \in H$. If the problem is changed to ensure $H$ is not the empty set, your proof is fine.