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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!

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  • $\begingroup$ it is known that $|U(n)|$ is the value Eulerphi(n), but Eulerphi(n) is always even $\endgroup$
    – janmarqz
    Commented Nov 13, 2015 at 0:10
  • $\begingroup$ Yes, I know - that is where I am running into trouble. Since I can't find a U group that is isomorphic, I am not quite sure how to proceed with this problem. $\endgroup$
    – pc724
    Commented Nov 13, 2015 at 0:15
  • $\begingroup$ the good news is that $73$ is prime : D $\endgroup$
    – janmarqz
    Commented Nov 13, 2015 at 0:15
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    $\begingroup$ @janmarqz $U_{73}$ is the wrong answer, though, since it is cyclic. $\endgroup$
    – Thomas Andrews
    Commented Nov 13, 2015 at 0:19
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    $\begingroup$ Plenty of candidates - you just need the nine to combine with either $4$ or $2$, which both can do. The smallest $5\cdot 3^3$ has the smallest prime divisors. $\endgroup$
    – Thomas Andrews
    Commented Nov 13, 2015 at 0:30

1 Answer 1

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$\mathbb{Z}_2\oplus \mathbb{Z}_4 \oplus \mathbb{Z}_9\cong\mathbb{Z}_2 \oplus \mathbb{Z}_{36}.$

$2+1$ 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.

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  • $\begingroup$ So the better problem would be find all $n$ instead of one. $\endgroup$
    – Groups
    Commented Nov 13, 2015 at 4:53

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