In the group $U_{1331}$, determine the number of elements of order 5.

Question

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!

Answer 1

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}

Answer 2

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 $\;\varphi(1331)=1210$.

Answer 3

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?

Answer 4

$x\in (\mathbb Z/1331\mathbb Z)^\times$ has order $5$ iff $x^5\equiv1\bmod1331$ and $x\not\equiv1\bmod1331$.

$x^5\equiv1\bmod1331=11^3\implies x^5\equiv1\bmod11$.

$2$ is a primitive root modulo $11$ because $2^5=32=33-1\equiv-1\bmod11$.

Therefore, the four residue classes $2^2=4$, $2^4=16\equiv5$, $2^6=64\equiv9$, and $2^8\equiv(2^4)^2\equiv25\equiv3$

have order $5\bmod11$.

We can "lift" these solutions to $\mod 121$; e.g., for the solution $4^5\equiv1\bmod11$ we have

$(4+11k)^5\equiv1\bmod121\implies4^5+5\times4^4\times11k\equiv1\bmod121$

$\implies11\times93+1+5\times4^4\times11k\equiv1\bmod121\implies93+5\times256k\equiv0\bmod11$

$\implies5(1+3k)\equiv0\bmod11\implies3k\equiv-1\bmod11\implies k\equiv7\bmod11.$

So $(4+77)^5=81^5\equiv1\bmod 121$.

Similarly, $5$ lifts to $27^5\equiv1\bmod121$, $9$ lifts to $9\bmod121$, and $3$ lifts to $3$.

(Alternatively, in retrospect, $3^5=243\equiv1\bmod 121$,

so $3^2=9$, $3^3=27$, and $3^4=81$ also have order $5$ mod $121$.)

Now these four solutions can be lifted to solutions $\bmod 1331$.

Take $x\equiv3\bmod121$ and set $(3+121k)^5\equiv1\bmod1331$.

Then $3^5+5\times3^4121k\equiv1\bmod1331\implies2+5\times3^4k\equiv0\bmod11$

$\implies2+5\times4k\equiv0\implies2+9k\equiv0\implies k\equiv1\bmod11$,

so $124^5\equiv1\bmod1331$.

Similarly, $9$ lifts to $735$, $27$ lifts to $632$, and $81$ lifts to $1170$.

These are the four elements of order $5$ in $(\mathbb Z/1331\mathbb Z)^\times.$