Find an integer n such that $U(n)$ is isomorphic to $\Bbb Z_2⊕\Bbb Z_4⊕\Bbb Z_9$ Find an integer n such that $U(n)$ is isomorphic to $Z_2⊕Z_4⊕Z_9$
I have gotten this far:
I know $Z_2$ is isomorphic to $U(4)$ and $Z_4$ is isomorphic to $U(5)$. However, I'm having trouble figuring out what $Z_9$ is isomorphic to in regards to the U-group. I remember proving somewhere that for all integers $n\geq 3$, $|U(n)|$ is even. Since $|Z_9|=9$, which is odd, I can't see to find a $U$ group that $Z_9$ is isomorphic to. 
Thanks for all the help! 

$\mathbb{Z}_2\oplus \mathbb{Z}_4 \oplus \mathbb{Z}_9\cong\mathbb{Z}_2 \oplus \mathbb{Z}_{36}.$
$2+1$ and $36+1$ are distinct primes, so $\mathbb{Z}_3\oplus \mathbb{Z}_{37}\cong \mathbb{Z}_{3.37}$ would work.
Similarly $\mathbb{Z}_2\oplus \mathbb{Z}_4\oplus \mathbb{Z}_9\cong \mathbb{Z}_4\oplus \mathbb{Z}_{18}$.
Then $4+1$ and $18+1$ are also distinct primes, so $\mathbb{Z}_5\oplus \mathbb{Z}_{19}\cong \mathbb{Z}_{5.19}$ would also work.
