Question

Let n be a positive even integer and h be a subgroup of zn of odd order

Answer 1

Answer:

3

Maybe try this: let G=Z/nZ≅ZnG=Z/nZ≅Zn. Then consider a subgroup HH of GG. Because GG is cyclic, HH is cyclic. Suppose HH does not consist of all even elements. Then there exists an odd element in HH. However, if HH is generated by an even element, then all elements in HH will be even. Therefore HH is generated by an odd element.

Now let's try to set up a bijection between even and odd elements of HH. Let H=⟨a⟩H=⟨a⟩ where aa is odd. Then (in additive notation) caca is even if cc is even, and odd if cc is odd.

Let ϕ(ca)=daϕ(ca)=da where d≡(c+1)(moda)d≡(c+1)(moda), 0≤d<|a|0≤d<|a|. This clearly maps between even and odd components. You can check that this is a bijection fairly easily.

There could be a much easier way to do this, but this is what I came up with.