Question

Let `G` be a group and `H` be a subgroup of`G=<a>` if`|G|=36`and`H=<a^4>`. How do you find `|H|` ?

Answer 1

`abs(H) = 9`

Explanation:

If I understand your notation correctly, `G` is a multiplicative group generated by one element, namely `a`.

Since it is also finite, of order `36` it can only be a cyclical group, isomorphic with `C_36`.

So `(a^4)^9 = a^36 = 1`.

Since `a^4` is of order `9`, the subgroup `H` generated by `a^4` is of order `9`.

That is:

`abs(H) = 9`