On the converse of Lagrange's theorem

Asked Mar 28 '20 at 07:42

Active Mar 28 '20 at 07:53

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We know that the order of an element in a finite group must divides the order of the group by Lagrange theorem. Is the converse true ? How do we prove it. I know that the converse is true for nilpotent groups.

edited Mar 28 '20 at 07:53

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asked Mar 28 '20 at 07:42

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There is no element of order $6$ in $;A_4;$ , which is of order $12$... – DonAntonio Mar 28 '20 at 07:44
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On the converse of Lagrange's theorem

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