## Characterization of parity in the left regular representation of a finite group

Let be a finite group and let be the left regular representation of . Prove that if is an element of order and if , then is a product of disjoint -cycles. Deduce that is an odd permutation if and only if is even and is odd.

This action of is faithful, so that the induced action of on is faithful. Note that under this action, for all we have , so that . Thus each -orbit of has order ; the -orbit of an element is precisely the cycle containing in the decomposition of since is cyclic with generator . Because , there are distinct orbits. Thus is a product of disjoint -cycles.

Clearly if is even and is odd, then is an odd permutation by Proposition 25.

Now if is an odd permutation and is odd, then the number of cycles in of even length is zero, which is even, so that is an even permutation- a contradiction. Thus is even. Now by Proposition 25, must be odd.

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## Comments

Faithfulness does not imply that the stabilizer of any point is . So the reasoning in the first paragraph is a little misleading. It will be better to say that:

This action of satisfies that the stabilizer of any point is the identity subgroup, so does the induced action.

Thank you; that was poorly worded.

typo: –

the cycle containing – the cycle containing

Thanks!