Prove that U(15) is not a cyclic group.
I know we can use the number of generators for a cyclic group to speed-up the proof.
$\endgroup$2 Answers
$\begingroup$By the Chinese remainder theorem we have that: $$ U_{15}\simeq U_3 \times U_5 $$ hence any element of $U_{15}$ has order $1,2$ or $4$. Since $|U_{15}|=\varphi(15)=8$, there is no element of $U_{15}$ that generates it, hence $U_{15}$ is not a cyclic group.
$\endgroup$ $\begingroup$Recall that cyclic groups have unique subgroups of each divisor order. So if $U(15)=\{1,2,4,7,8,11,13,14\}$ were cyclic, it would have exactly ONE subgroup of order 1, order 2, order 4, and order 8.
This then implies that it would only have ONE element of order 2 (since each element of order 2 generates a distinct subgroup of order 2).
But notice that $14^2=1$ and $11^2=1$ so both $14$ and $11$ have order 2. Thus $U(15)$ must not be cyclic.
Alternatively, you could just compute the order of each element and see that it has no elements of order 8.
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