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I'm sorry if I am misunderstood. My native language is not English and this is my first time using this site.

I have home-work, to prove or to disprove the following statement:

Let $m,n$ be natural numbers such that $g.c.d(m,n) = 1$. Then:

The direct sum of the multiplicative groups $U_n$ and $U_m$ (where $U_n$ is the multiplicative group that contain the numbers that are relatively prime with $n$,), that is, the group $(U_n \times U_m)$ is a multiplicative group and isomorphic to $U_{mn}$

I've been trying to solve this for quite long time and I wasn't very successful with that.

I've noticed that $|U_{mn}| = |U_m||U_n|$ so with the obvious homomorphism $f(a) = (a \pmod m), a\pmod n)$ I just need to show that the function is either surjective or injective in order to show that the function is bijective and then it is a homomorphism.

I was trying to show that the kernel is empty but wasn't successful with that. I was also trying to show that for every element in the image there exist element in the domain that mapped to him but also I wasn't successful with that.

Can anyone help me?

hadadrefael
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2 Answers2

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Here is the standard roadmap for that result:

  • Use the Chinese remainder theorem to get $\Bbb Z/mn\Bbb Z \cong \Bbb Z/m\Bbb Z \times \Bbb Z/n\Bbb Z$ as rings.
    An explicit isomorphism is exactly your $f$.

  • Let $U_n=(\Bbb Z/n\Bbb Z)^\times$ and get $U_{mn} \cong U_{m} \times U_{n}$ from this general result:

If $A$ and $B$ are commutative rings with unity, then $(A \times B)^\times = A^\times \times B^\times$, where $R^\times$ is the group of units of the ring $R$.

lhf
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Hint:

The three fit together due to coprimality.

NB: The notation $(\Bbb Z/n\Bbb Z)^\times=U_n$ is standard.

Shaun
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