An abelian group proof with $g*g=e$ for all $g$.

Question

I have to show that the following group $$ (G, * , e) $$ with its operation $*$, which is defined through $ g*g = e$ for every $g \in G $ is an abelian group.

In order to do that one have only to show that the group is commutative.

How can one prove it whereas the operation is defined always between an Element and itself?

I reckon it is not so simple as it seems

Thanks in advance for your help :)

Answer 1

Consider $$(a\ast b)\ast (a\ast b)=e=(a\ast a)\ast(b\ast b)$$ then cancel $a$ on the left and $b$ on the right.

Answer 2

to answer your actual question.

When the book says $*$ is defined to be that $g*g=e$ for all $g\in G$, the book is abusing language. That isn't the definition of $*$; it is a property (a very significant property) of $*$ but not the definition of $*$.

If $G$, the set, is $\{e,a,b,c,d,e,f,g\}$ for example we know that $a*a=e$ and $c*c = e$ but we don't know what $ab$ is equal to; It could by $ab = c$ or it could be $ab = f$. But it doesn't matter.

......

We can prove that for all $a,b \in G$ that whatever $ab$ is equal to, that $ba$ must also be equal to the same value.

Note that if $ab = k$ and $k*k = e$ we have $(ab)(ab) = e$. But we also have $(ab)(ba) = a(bb)a = aea =aa = e$. So we have both $(ab)(ab) = e$ and $(ab)(ba) = e$.

If we rely on the proof we should have proven earlier that inverses are unique that means that $ab = (ab)^{-1}$ and $ba = (ab)^{-1}$ so $ab = ba$.

.....

If we want to really convince ourselves and we don't mind reinventing the wheel:

$(ab)(ab) = e$

$(ab)(ab)(ba) = (ba)$

$(ab)a(bb)a = ba$

$(ab)aa = ba$

$(ab) = ba$

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The thing is that other than examples to teach the concept of groups the actual operation $*$ doesn't matter and we almost never define what $a*b$ is equal to. And if $ab = c$ or $ab=f$ it won't matter. That's just labelling. What matters is only certain properties.