# Any permutation can be only either even or odd. [duplicate]

A permutation can't be both even and Odd. How?? Is their any proof?? Kindly tell me.!

Thanks beforehand

## marked as duplicate by Martin R, Zain Patel, JMP, Namaste abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 2 '17 at 12:53

Given a permutation $\pi$ we define $$\sigma(\pi) = |\{(i, j) : i < j \textrm{ and } \pi(i) > \pi(j)\}| \mod 2$$
It is easy to show inductively that $\sigma$ is $1\mod 2$ iff $\pi$ can be expressed as the product of a odd number of transpositions. By showing every transposition changes $\sigma$ by one. Once you have convinced yourself of this it follows that the parity of permutation is well defined.