Question

prove that in a group of even elements at least one element apart from the identity element must equal to its identity
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Answer 1

SOLUTION

TO PROVE

In a group of even number of elements, at least one element, apart from the identity element, must equal to its identity

PROOF

Let G be the group

Let a ∈ G

$\sf{Then \: \: o(a) = o( {a}^{ - 1} )}$

$\sf{If \: \: o(a) < 3 \: \: then \: \: a = {a}^{ - 1} }$

$\sf{If \: \: o(a) \geqslant 3 \: \: \: then \: \: a \: \: and \: \: {a}^{ - 1} \: \: distinct }$

and they form a pair of elements of G such that Each is the inverse of the other.

Let us consider all pairs of elements of the form

$\sf{\: \: \{ \: x, {x}^{ - 1} \} \: } \: where \: \: o(x) \geqslant 3$

All such pairs can not exhaust the whole G, because G contains atleast one element ( the identity element) of order less than 3

Since the order of G is even, an even number of elements having order less than 3 lie outside the union of all pairs of the form

$\sf{\: \: \{ \: x, {x}^{ - 1} \} \: } \: where \: \: o(x) \geqslant 3$

Since G contains only one element of order 1, the number of elements of order 2 must be odd

Hence in particular G must contain at least one element of order 2

Hence the proof follows

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