Find total number of elements of order 20 in the multiplicative group $\mathbb Z^*_{100}$ How can I find all the elements of order $20$ in the multiplicative group $\mathbb Z^*_{100}$.   
$[7]\in \mathbb Z^*_{100}$.
$7^4\equiv 1\pmod{100}$. So order of $[7]=4$. But how can I find all those elements of order $20$ in $ \mathbb Z^*_{100}$?.
Are there any theorem of formula to find order of an element?
Than you in advance.
 A: By chinese remainder theorem $\mathbb Z_{100}^*\cong \mathbb Z_4^*\times \mathbb Z_{25}^*$.
An element of the latter subgroup $(a,b)$ has order $lcm(|a|,|b|)$ where $|a|$ and $|b|$ are the orders in $\mathbb Z_4^*$ and $\mathbb Z_{25}^*$. So if $(|a|,|b|)=20$ we must have $|b|=20$.
Of course it is a well known theorem $\mathbb Z_{p^\alpha}^*$ is cyclic for $p$ an odd prime.
So the pair $(a,b)$ has order $20$ if and only if $|b|=20$. Since $\mathbb Z_{25}^*\cong \mathbb Z_{20}$ there are $\varphi(20)=8$ suitable options for $b$ (cyclic groups have $\varphi(n)$ elements of order $n$)and hence $2$ options for $a$ ($\mathbb Z_{4}^*$ has two elements).
Hence there are $16$ elements of order $20$.
If you want to find them all then one way to do it is by finding a generator for $\mathbb Z_{25}^*$ 
Lets try with $2$, we get $2,4,8,16,7,14,3,6,12,24,23,21,17,9,18,11,22,19,13,1$
So in fact $2$ generates. From here we can see the groups as:
$2,2^2,2^3,2^4,2^5\dots 2^{20}$, of course we want the elements in which the exponent is relatively prime with $20$. So we want $2^{1},2^3,2^7,2^9,2^{11},2^{13},2^{17},2^{19}$ which corresponds to $2,8,3,12,17,22,13$.
Each of these values for $b$ shall give us two congruence classes $\bmod 100$ of order $20$, they are $(1,b)$ and $(3,b)$ respectively. It is clear these numbers are $b,b+50$ if $b$ is odd and $b+25,b+75$ if $b$ is even. Doing this for each $b$ is all that remains, lets do it:
$27,77$ results from $b=2$ and $33,83$ result from $b=8$. The other $12$ are:
$3 ,53, 37, 87, 23, 73, 17, 67, 47, 97, 13, 63$. So these are the $16$ elements.
