Is U-group isomorphic to $\mathbb{Z}_4 \oplus\mathbb{Z}_4$?

Question

How can i show that no $U(n)$ is isomorphic to $\mathbb{Z}_4 \oplus\mathbb{Z}_4$? Please give me some hint. Thanks in advance.

Let $U(n)$ denote $\Bbb Z_n\oplus\Bbb Z_n$. Then your statement is false. — Berci, Mar 12 '13 at 17:29
I think that he means $U_n$ the group of $n^{th}$ roots of the unity. — , Mar 12 '13 at 17:32
I think the OP means the group of units of $\mathbb{Z}/n\mathbb{Z}.$ — mv3, Mar 12 '13 at 17:32

Andreas Caranti · Accepted Answer · 2013-03-12T18:10:18.657

3

I take $U(n)$ to mean the group of invertible elements in $\Bbb{Z}_{n}$.

If $$ n = \prod p_{i}^{e_{i}} $$ is the decomposition on $n$ as the product of powers of distinct primes, then $$ U(n) \cong \oplus U(p_{i}^{e_{i}}). $$

So we ask first (thanks @AlexYoucis) whether we can have $$ U(p_{i}^{e_{i}}) \cong \Bbb{Z}_4 \times \Bbb{Z}_4. $$ By order reasons, this could only happen for $p_{i}^{e_{i}} = 17$ (but then $U(p_{i}^{e_{i}}) \cong \Bbb{Z}_{16}$ is cyclic of order $16$), or for $p_{i}^{e_{i}} = 32$, as $\varphi(32) = 16$. But $U(2^k) \cong \Bbb{Z}_{2^{k-2}} \times C_2$ is known to be non-cyclic for $k \ge 3$.

Then we ask when we can have $$ U(p_{i}^{e_{i}}) \cong \Bbb{Z}_4. $$

This can only happen when $p_{i}^{e_{i}} = 5$, because $U(8) \cong \Bbb{Z}_2 \times \Bbb{Z}_2$. And it is easy to see that $\varphi(n) = 4$ only for these two values.

So, indeed no $U(n)$ is isomorphic to $\Bbb{Z}_4 \times \Bbb{Z}_4$.

edited Mar 12 '13 at 18:10

answered Mar 12 '13 at 17:36

Andreas Caranti

68,873

@AlexYoucis, you're right, thanks, fixing now. – Andreas Caranti Mar 12 '13 at 17:41
You're welcome. :) – Alex Youcis Mar 12 '13 at 17:47
i can't understand because here we want to show that U(n) is not isomorphic to$ \mathbb{Z_4} \oplus \mathbb{Z_4}$ – Siddhant Trivedi Mar 12 '13 at 17:56
@SiddhantTrivedi, you want to show that $U(n)$ is not isomorphic to the direct sum of two cyclic groups of order $4$. I just called them $C_4$ and not $\Bbb{Z}4$ not to mix up things with the units of $\Bbb{Z}{n}$, but it's exactly your problem. – Andreas Caranti Mar 12 '13 at 18:01
@SiddhantTrivedi, but I'll switch to your notation now, since it's the same things. – Andreas Caranti Mar 12 '13 at 18:08
ok i understand – Siddhant Trivedi Mar 12 '13 at 18:10

Julien · Answer 2 · 2013-03-12T22:11:27.130

Here is a slightly silly approach. Assume $U(n)$ is isomorphic to $\mathbb{Z}_4\oplus \mathbb{Z}_4$. This forces $\phi(n)=16$. One can play around with this and the formula $$\phi(p_1^{r_1}\cdots p_s^{r_s})=(p_1^{r_1}-p_1^{r_1-1})\cdots (p_s^{r_s}-p_s^{r_s-1})=p_1^{r_1-1}\cdots p_s^{r_s-1}(p_1-1)\cdots (p_s-1)$$ to find that the only such $n$'s are $$ 17,\;32,\;34, \;40,\;48,\; 60. $$ By the chinese remainder theorem $U(k\cdot l)\simeq U(k)\oplus U(l)$ whenever $k$ and $l$ are relatively prime. For $p$ prime $U(p)\simeq \mathbb{Z}_{p-1}$. Besides, due to the presence of $-1$ (order $2$) and $5$ (order $2^{r-2}$) whose generated subgroups have trivial intersection, $$ U(2^r)\simeq \mathbb{Z}_2\oplus \mathbb{Z}_{2^{r-2}}. $$

It follows that $17$ and $34=2\cdot 17$ yield $\mathbb{Z}_{16}$. Impossible.

Then, $40=2^3\cdot 5$, $48=2^3\cdot 3$, and $60=2^2\cdot 3\cdot 5$ yield three nontrivial $\mathbb{Z}_{2^j}$ summands. Still impossible.

And finally $32=2^5$ gives $\mathbb{Z}_2\oplus \mathbb{Z}_8$. It does not work either.

@TobiasKildetoft Hum.. I took it was $\mathbb{Z}/n\mathbb{Z}$... I should have asked... — Julien, Mar 12 '13 at 17:32

Is U-group isomorphic to $\mathbb{Z}_4 \oplus\mathbb{Z}_4$?

2 Answers2