Clarification regarding a group theory proof In a group we have $abc = cba$. If $c \neq 1$, is the group abelian? See the following link. 
(I am new to this site but it is my understanding that you cannot PM authors, correct? Which is too bad because it means I have to open this thread)
In regards to Math Gem's answer, from the comments I take it he has omitted some of the obvious statements; but I wonder, if one was to give every required statement, what would have to be included?
In the proof he requires $c = ba$ which surely is only a small (equal to the order of the group times the number of $c \neq 1$) section of a group. It must hold for all $a,b, c \neq 1$ so the total amount of combinations you would have to verify (say A is the group) would be $\mathrm{T} = \#A \cdot \#A \cdot (\#A-1)$. Say $\#A = x$. Since $c = ba$ is only valid for $x$ distinct couples $(a,b)$ (because of the "latin square property", $c = ba$ "misses" $\mathrm{T}-(x-1)\cdot\#A$ combinations, right? 
Taking $Z_4,+$ and $2$ for example, only $2 = 2 + 0$, $2 = 0 + 2$, $2 = 1 + 3$ and $2 = 3 + 1$ so e.g. (4,4,4) and (1,2,4) are invalid for the proof given, yet we do have to take them into account. From the comments, if $ab = 1$ then clearly $ab = ba$ (because $ab = 1 \implies a = b^{-1} \implies 1 = ba$) but then there's still e.g. (4,4,4) and (1,2,4). So the trivial (4,4,4) would have to be mentioned, as well when either $a$ or $b$ is equal to $1$. From inspection of $Z_4,+$ I see that we now have all possible combinations, but should it not be proven? I.e. how do you know there do not exist $a,b, c$ such that $ba \neq c$ and $a,b \neq 1$ and yet $ab \neq ba$?
For $Z_4,+$ you can sort of say: the order is 4, so you have to check $4\cdot4\cdot3$ combinations. The proof holds for $c = ba$ and since there 3 elements in $Z_4,+$ that are not equal to $0$, you substract $4\cdot3$. Then you remove (1,b,c) and (a,1,c) so you substract $4\cdot3$ twice. Then for $ab = 1$ you substract $4\cdot3$ again and you are left with $0$. So yeah, it's true for $Z_4$, but it's not exactly a real proof. Maybe there is some peculiar group out there for which there are more elements left?
I guess you would just generalize it to some order $x$ but then I don't get $0$ like I did above.
So my  question is how do you know you have all possible combinations $a,b, c, \neq 1$ when basically your proof consists of checking different cases.
(I am a 1st year undergraduate math student, should it matter)
 A: He doesn't require $c=ba$; he just fixes any $a,b$, and then applies the equality (which holds for any $c$, so in particular for any one you might like to consider) $$abc=cba$$ to the special case where $c$ is taken to be $ba$. This is then sufficient to prove that $$ab=ba$$ since you get $abba = baba$ and you can then multiply successively by $a^{-1}$ then $b^{-1}$ on the right of both sides. Since $a$, $b$ were arbitrary, this proves that $ab=ba$ for all elements $a$, $b$, i.e. that the group is abelian.
A: If $abc=cba$ for any $c$ whatever, let $c=ba$. Then we have $$abba=baba$$
So that $$(abba)a^{-1}b^{-1}=(baba)a^{-1}b^{-1}$$ which reduces to $$ab=ba$$
A: He doesn't require $c=ba$, he (and that is justified) uses a special case of the given condition.
What is needed to show is
$$\tag1 \forall x\forall y\colon xy=yx$$
and we are given
$$\tag2\forall a\ne1\forall b\ne1\forall c\ne1\colon  abc=cba$$
We do not have to check these $(n-1)^3$ equalities (2), we are given that they hold. If we consider some of the instances supoerfluous, we are free to simply ignore them.
On the contrary, we have to show the $n^2$ equalities (1).
Notably, the case $x=1$ in (1) holds for any group by the properties of the neutral element. Hence we may assume that $x\ne 1$.
Likewise, (1) trivially holds when $y=1$, hence we may assume that $y\ne 1$.
And finally each element commutes with its inverse, hence the case $y=x^{-1}$ is also automatically true in (1).
So we can assume $x\ne1$, $y\ne 1$, $yx\ne 1$. Thus allows us to specialize  (2) to (letting $a=x, b=y, c=yx$)
$$xyyx=yxyx$$
which implies (by cancelling from the right)  $$xy=yx$$ as was to be shown.
