# decomposing a factor group into a direct sum of cyclic groups

Let $\mathbb{Z}_n ^*$ be the set of all units in $\mathbb{Z}_n$ and $(\mathbb{Z}_n ^*)^2$ = $\{ a^2 | a \in \mathbb{Z}_n ^*\}$

Then, decompose the factor group $\mathbb{Z}_n ^* / (\mathbb{Z}_n ^*)^2$

1.when $n$ is an odd prime

2.when $n$ is a product of two distinct odd primes $p, q$

I solved the first as following: In $\mathbb{Z}_p$, since $x^2=(p-x)^2$ holds, the order of $(\mathbb{Z}_n ^*)^2$ is exactly the half of that of $\mathbb{Z}_n ^*$.

Thus, the answer is $\mathbb{Z}_2$.

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Hints:

$$\Bbb Z_{pq}^*=\Bbb Z^*_p\times\Bbb Z^*_q$$

The above's based on the well known property of Euler's Totient Function

$$\phi(pq)=\phi(p)\phi(q)=(p-1)(q-1)$$

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I know that both groups have the same order. However, since both p-1 and q-1 may not be prime, may your hint be incorrect? For example, if p=3, q=5, the group in the left has order 8, which is the order of $\mathbb{Z}_8$, different from $\mathbb{Z}_2$ X $\mathbb{Z}_4$, – NH.Jeong Oct 14 '13 at 15:44
The group $\;\Bbb Z_{15}^*\;$ is not cyclic...! – DonAntonio Oct 14 '13 at 15:55
Oh I checked that they are isonorphic. You are right. Then is the answer of the second problem is the klein 4group because $\mathbb{Z}_{pq}^*$ is not cyclic? – NH.Jeong Oct 14 '13 at 16:23
How can that be the Klein group if $\;p,q\;$ different odd primes?? – DonAntonio Oct 14 '13 at 16:26
Um.. I applied the argument in my question, so I thought the order of $(\Bbb Z^*_p\times\Bbb Z^*_q)^2$ is the 1/4 times of that of $\Bbb Z^*_p\times\Bbb Z^*_q$ – NH.Jeong Oct 14 '13 at 16:34

$\mathbb{Z}_{pq}^\ast = \mathbb{Z}_p^\ast \times \mathbb{Z}_q^\ast$, and $(\mathbb{Z}_{pq}^\ast)^2 = (\mathbb{Z}_p^\ast)^2 \times (\mathbb{Z}_q^\ast)^2$.

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Hence is the answer is klein group? – NH.Jeong Oct 16 '13 at 8:17
@JeongNam-ho That should be easy enough to check, no? – Slade Oct 16 '13 at 10:00
What are you saying..?T.T. So, Is it the klein group?? – NH.Jeong Oct 16 '13 at 11:16
@JeongNam-ho Well, what is the answer for $n=15$? How would a confident mathematician ask this question? – Slade Oct 16 '13 at 11:27
I think, the answer is the klein group for the case n=15..... – NH.Jeong Oct 16 '13 at 12:23