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I know that there are $gcd(a,b)$ unique group homomorphisms between $\mathbb{Z}/a\mathbb{Z}$ and $\mathbb{Z}/b\mathbb{Z}$ for $a,b \in \mathbb{N}, a,b \geq 2$. I just want to prove it in a simple way.

A group homomorphism $\phi$ of the desired kind is uniquely defined by the choice of $\phi(\bar{1})$.

Let $X$ be the number of possible group homomorphisms. I want to show that $X$ divides $a$ and $b$, and is therefore exactly $gcd(a,b)$.

Assumption: $\phi(\bar{1}) \neq \bar{0}$ (the trivial case).

For a group homomorphism $\phi(a) + \phi(b) = \phi(a + b)$ has to apply. Therefore $b = |\phi(\bar{1})| \cdot k$ for some $k \in \mathbb{N} \Rightarrow |\phi(\bar{1})|$ divides $b$. This is required in order to express the neutral element as a sum of two other elements (and achieve the "wrap").

I just discarded my approach why $|\phi(\bar{1})|$ needs to divide $a$ as well as it was nonsense. It seems logical to me that this needs to apply, but I don't have an idea how to show it.

Can you help me to finish this? Is the first part correct? How would I describe it in a more formal way?

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  • $\begingroup$ Careful. $1$ divides $6$ and $9$ but certainly isn't $\gcd(6,9)$. $\endgroup$ – user98602 Nov 6 '14 at 22:17
  • $\begingroup$ What do you mean by $|\phi(\bar{1})|$? $\endgroup$ – user44400 Nov 6 '14 at 22:19
  • $\begingroup$ @user44400 the order of $\phi(\bar{1})$, so the smallest integer x with $\left(\phi(\bar{1})\right)^x = e$ $\endgroup$ – muffel Nov 6 '14 at 22:22
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You made a good start. What you have is that $a\phi(1) =\phi(a) = \phi(0)= 0$.

Now note that this means that $a$ is a multiple of the order of $\phi(1)$, that is the order is a divisor of $a$.

But, the order of each element in $\mathbb{Z}/b \mathbb{Z}$ is a divisor of $b$.

So, you get that the order of $\phi(1)$ divides $a$ and $b$, and you get the result soon.

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  • $\begingroup$ thank you for your answer! Just one question: can you please explain why the order of each element in $\mathbb{Z}/b\mathbb{Z}$ is a divisor of $b$? $\endgroup$ – muffel Nov 6 '14 at 23:05
  • $\begingroup$ You are welcome. For your question. It is just like with the $a$. Given some element $c$ you certainly have $bc=0$ in $Z/bZ$. So $b$ is a multiple of the order of $c$, that is the order is a divsor of $b$. $\endgroup$ – quid Nov 6 '14 at 23:10
  • $\begingroup$ Thanks! But if I take $\mathbb{Z}/4\mathbb{Z} = \{0,1,2,3\}$, the order of $2$ is $3$ which is not a divider of 4. Am I missing something? $\endgroup$ – muffel Nov 6 '14 at 23:37
  • $\begingroup$ But 2 times 2 is 4 which is 0 so the order of 2 is 2. I am not sure which reasoning leads you to the order of 2 being 3. One thing you should be careful about: we are in a group here that is written additively (not multipilcatively as is common for groups too) so g has order d means dg = 0. $\endgroup$ – quid Nov 7 '14 at 10:59
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    $\begingroup$ my fault, sorry, now I got it. Thank you very much! $\endgroup$ – muffel Nov 7 '14 at 14:42

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