In a finite group we have $abc\cdots z = z\cdots cba$ . Is it abelian?

Question

In a group we have $abc\cdots z = z\cdots cba$ then is it an abelian group? It's proven for $3$ members here:

In a group we have $abc=cba$. Is it abelian?

And I do not find a way to try an inductive proof (If it's right for three members then it's right for n members).

In case this property holds for all choices of $a,b,...,z$ just take $c=d=...=z=1$, where $1$ denotes the neutral element of the group. — Jonas Linssen, Oct 25 '20 at 13:36
If $a,b,c,\dots,y, z$ are arbitrary, let each of $b,c,\dots, y$ be the identity; then $az=za$ holds for arbitrary $a,z$. — Shaun, Oct 25 '20 at 13:37
Looking at the linked question, I think it should be assumed that all elements $a,\ldots, z$ are $\ne 0$. Otherwise it’s trivial, as the other comments point out — Milten, Oct 25 '20 at 13:42

1123581321 · Answer 1 · 2020-10-25T13:52:34.830

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Following the question in the link I'll try to show the case where:

$abcd=dcba$ for all $a,b,c,d\not=1$ in $G$. It is $a'=a^{-1}$ and

\begin{align}abcd=dcba \iff \\ abcda'b'c'd'=1 \\ \iff abcd(cba)'d'=1 , \\ \ \text{ for } d=cba \\ abc(cba)'=1 \iff \\ abca'b'c'=1 \iff \\ abc(ba)'c'=1, \\ \text{ for } c=ba \\ aba'b'=1 \checkmark \end{align}

edited Oct 25 '20 at 13:52

answered Oct 25 '20 at 13:45

1123581321

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This is difficult to read. Please use the "align" $\LaTeX$ feature. – Shaun Oct 25 '20 at 13:51
It's . . . still unclear. I don't know how you managed that. – Shaun Oct 25 '20 at 17:17

In a finite group we have $abc\cdots z = z\cdots cba$ . Is it abelian?

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