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This is the same question as was asked about in this topic, but I can't post in that one (presumably because it's archived?)

1. The problem statement, all variables and given/known data

Suppose G is a group with |G| = 4n+2. Show that there is a subgroup H < G such that |H| = 2n + 1. Use Cauchy's theorem, Cayley's theorem and the fact that any subgroup of [itex]S_n[/itex] has either all of its elements or precisely half of its elements being even.

2. Relevant equations

Cayley: Every group G is a subgroup of [itex]S_|_G_|[/itex].

Cauchy: If G is a finite group & P [tex]\in[/tex] N is prime with P | |G|, then G contains an element of order P.

3. The attempt at a solution

Cayley [tex]\Rightarrow G \leq[/tex] [itex]S_4_n_+_2[/itex] [tex]\Rightarrow[/tex] all or half of its elements are even (*)

Cauchy [tex]\Rightarrow[/tex] G contains an element of order 2 (since 2 | 4n+2)

[tex]\Rightarrow[/tex] G contains an element that's a product of disjoint transpositions.

Suppose G contains an element that's a product of anoddnumber of disjoint transpositions. (**)

[tex]\Rightarrow[/tex] Half the elements in G are even, by (*), i.e. 2n+1 of them are even.

These 2n+1 elements form a group (since they're even, their products are even, their inverses are even, and 1 is even), which is a subgroup of G with order 2n+1.

... So what I need to do now is show that G contains an element that's a product of an odd number of disjoint transpositions, i.e. prove that (**) is true.

And that's where I'm stuck... I have no idea what to do.. Can I have a nudge in the right direction?

Thanks

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# Group theory hints?

Can you offer guidance or do you also need help?

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