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I think it's not true. But since in a group, inverse for each element exist then, we can write:

$ab=e$
Multiplying by $a^{-1}$ on both sides
$b=a^{-1}e=a^{-1}$

This implies that $ba=e$

Am I missing out something? Thankyou for any help or hint.

Esha
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    If you multiply $ab=e$ on the left by $a^{-1}$ (not "both sides") you get $b=a^{-1}$. If you then multiply $b=a^{-1}$ by $a$ on the right you get $ba=e$. – anon Apr 05 '21 at 05:06
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    $ba=bae=babb^{-1}=beb^{-1}=bb^{-1}=e$ – bof Apr 05 '21 at 05:11
  • This is true if $a$ and $b$ are elements of a group and $e$ denotes the identity element of that group. Perhaps you should have said that rather than leaving it to the reader to guess what you are talking about? – Derek Holt Apr 05 '21 at 07:54
  • In general, for every $a$ and $b$ in a group, we get $ba=b(ab)b^{-1}$ (which boils down to $ba=ab$ if the group is abelian). So, what does heppen if, for some $a,b$ in the group, $ab=e$? –  Apr 05 '21 at 08:10

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