Hey so i tried this but i dont really know if its correct. I need some suggestions.
Let $G$ be $\{e,a,b,c\}$ Let's assume $G$ is not abelian.
This means there exist $a,b$ such that $ab\neq ba$.
We can choose 2 elements $a, b$ who are not inverse to each other.
We know that $(ab≠b \land ab≠e \land ab≠a) \implies ab=c$; we also know that $(ba≠b \land ba≠e \land ba≠a \land ab≠ba) \implies ba=d$ which is a contradiction to our assumption that G had 4 elements.
Is this proof legit ?
Your proof is good, but it can be made more elegant. This is a matter of experience... keep practising!
Let G be {e,a,b,c} Let's assume G is not abelian.
This means there exist a,b such that ab≠ba.
It's strange to say "there exist $a,b$" when $a$ and $b$ are already explicit objects. Better wording would be, "there exist two elements, say $a$ and $b$, that do not commute." Alternatively, "without loss of generality, assume $ab \neq ba$."
We can choose 2 elements a, b who are not inverse to each other.
Again it is strange to be "choosing" $a$ and $b$ at this point since we are already working with them. Better: "Note that $a$ and $b$ are not inverses of each other, for then $ab=ba=e$."
We know that (ab≠b ^ ab≠e ^ ab≠a)-> ab=c;
You could write this more smoothly thus: "Since $ab \notin \{a,b,e\}$, it follows that $ab = c$."
... we also know that (ba≠b ^ ba≠e ^ ba≠a ^ ab≠ba) -> ba=d which is a contradiction to our assumption that G had 4 elements.
I understand what you mean here, but what exactly is $d$? If you really want to keep this $d$, it would be clearer to write, "$ba = d$ for some $d \notin G$, contradicting the closure of $G$". But this is more complex than necessary; avoid introducing new objects if you can. Instead, just appeal to symmetry: "By the same reasoning, we also have $ba = c$. But then $ab = ba$, a contradiction."
The revised proof:
Let $G = \{e,a,b,c\}$. Let's assume $G$ is not abelian; without loss of generality, assume $ab \neq ba$. Note that $a$ and $b$ are not inverses of each other, for then $ab = ba = e$.
Since $ab \notin \{a,b,e\}$, it follows that $ab=c$. By the same reasoning, $ba=c$. But then $ab=ba$, a contradiction.
If $a\in G$ has order 4. done, if not there exists $a\neq b$ of order 2 and $a\neq b^{-1}$. We have $ab\neq a$ otherwise $a^{-1}(ab)=1=b$, you also have $ab\neq b$. The group is $\{1,a,b,ab\}$, but $ba\neq a, ba\neq b$, so $ba=ab$.
The order of each element divides the order of the group. If there is an element of order $4$, then $G$ is cyclic and trivially abelian.
If there is no element of order $4$, then every (non-identity) element has order $2$. Can you show the following general proposition?
Lemma: If every element of a group $G$ has order $2$, then $G$ is abelian.
There are two cases.
If $G$ is cyclic, then $G$ is obviously commutative.
If $G$ is not cyclic, then every non-trivial element has order $2$. Let then $e \ne x \in G$: we then have the subgroup $\{ e, x \}$. Pick now $y \in G \setminus \{ e, x \}$, therefore so far we have the elements $\{ e, x, y \}$. What about the elements $xy$ and $yx$? Well, since $G$ has only $4$ elements, it follows that some two of $\{ e, x, y, xy, yx \}$ must be equal. Since $x, y \ne e$, the only possibility is $xy = yx$, which means commutativity.
