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

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Answer 1 · 2018-01-18T19:59:07

$abs(H) = 9$

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:

$abs(H) = 9$

Let GG be a group and HH be a subgroup ofG=<a>G=<a> if|G|=36|G|=36andH=<a^4>H=<a^4>. How do you find |H||H| ?