In the group $U_{1331}$, determine the number of elements of order 5. Please advise whether my solution is correct:
Know: 
$1331 = 11{^3}$ -> Have: $\phi$(11) = 10 -> and 5 | 10
So can use the following theorem: If p is an odd prime and d | (p - 1) . (where d = order of elt. a $\epsilon U_{p}$) Then the number of elements $\epsilon U_{p}$ of order d is $\phi(d)$
In this case: $\phi$(5) = 4. So there are 4 elts. of order 5?
Is this true? Since $1331 = 11{^3}$ -> $\phi$(1331) = $\phi(11)\phi(11)\phi(11)$ = 30 = # of primitive roots?
Also would there be a short cut to determining all these elts. of order 5?
Thanks!
 A: First, $\phi$ is not fully multiplicative; in particular, it is. not true that $\phi(11^3)=\phi(11)^3$ (and also, $10\times10\times10=1000$, not $30$, but that is irrelevant). The function $\phi$ is multiplicative: if $\gcd(a,b)=1$, then $\phi(ab)=\phi(a)\phi(b)$; but not fully multiplicative.
Second, for $p$ an odd prime, the group $U(p^n)$ of units modulo $p^n$ is known to be cyclic. If you know this fact, then you can use it, because in a cyclic group, there is one and only one subgroup of order $d$ for any $d$ that divides the order. So there is one and only one subgroup of order $5$ in $U(11^3)$ if and only if $5|\phi(11^3)$ (otherwise, there are none). And how many elements of order $5$ are there, then?
A: There are two errors.
First, there  can be no element of order $5$ in a group of order $1331$, since $5$ is not a divisor of $1331$ (think of Lagrange's theorem).
Second, for any prime $p$, and any $n>0$, $\;\varphi(p^n)=p^{n-1}(p-1)$, so $\;\varphi(1331)=1210$.
A: You might find this article helpful.
Since $\varphi(1331) = \varphi(11^3) = 11^3-11^2 = 1331 - 121 = 1210$, it follows that $|U_{1331}|=1210$.
Acording to Wolfram alpha
$$2^{\text{Divisors}[1210]} \pmod{1331}
  =\{2, 4, 32, 1024, 717, 323, 362, 606, 596, 1170, 1330, 1\}$$
It follows that $U_{1331}=\langle \bar 2 \rangle$
Hence there are $\varphi(5) = 4$ elements of  $U_{1331}$ of order $5$.
Those elements would be 
\begin{align}
   2^{1\cdot 242} \mod{1331} &= 1170 \\
   2^{2\cdot 242} \mod{1331} &= 632 \\
   2^{3\cdot 242} \mod{1331} &= 735 \\
  2^{4\cdot 242} \mod{1331} &= 124
\end{align}
