Number theory related to crypto I have a question related to a piece of coursework that is for cryptography and more for encryption that relies on Number theory, now I have no knowledge of number theory and the tutor did not cover it well enough, and I am starting to learn it slowly, but I have an exam coming up and one question will be like the one below, so any help from you guys would be appreciated.enter image description here
Consider the group $G=\mathbb{Z}^*_{61}$ with respect to multiplication. Compute the following
o Find the inverse element of 53 in $G$.
o Compute the order of the element 15 in $G$.
o Let $H := \langle 8 \rangle$ be a group containt in $G$, i.e. H is generated by the of powers of 8 modulo 61. Find two non-trivial subgroups of H. The trivial subgroups of $H$ are $\langle 1 \rangle$ (which only contains the identity element and $H$ itself.

Since p (here p=61) is a prime you should know that it holds $x^{p-1} = 1 \mod p$.
Therefore $x^{p-1-u} \cdot x^u = 1 \mod p$ and so $x^{p-1-u}$ and $x^u$ are inverses. Therefore $x^{-1} = x^{p-2} \mod p$  
To answer the first point we simply have to calculate $53^{59} \mod 61$, which is 38. You can check this by, $53 \cdot 38 \equiv 1 \mod 61$   
To compute the order of an element x, we simply have to find the lowest power n, where $x^n \equiv 1 \mod p$ (note that n has to be a divisor of p-1). In our case we find that $15^{15} \equiv 1 \mod 61$ and therefore order of 15 is 15.  
Note that H has 20 elements since the order of 8 is 20. To find subgroups of H, you have to find subgroups of order of divisors of 20 (2,4,5,10) and therefore simply solve the equations $x^y \equiv 1 \mod 61$, where $y \in \{2,4,5,10\}$. To find a subgroup with just 2 elements we can take $\langle 60 \rangle$, which only contains the elements 60 and 1. For a subgroup with 4 elements we can take $\langle 50 \rangle$ or $\langle 11 \rangle$, which contain the elements 50,60,11,1.
