# Why is $\phi(x^i)=y^i$ not a group homomorphism between the cyclic group of order $36$ to the cyclic group of order $17$?

In Artin Algebra 2.10.3 there gives a group homomorphism from a cyclic group of order $12$ to a cyclic group of order $6$. Defined by $\phi(x^i)=y^i$ with $x$ in the cyclic group of order $12$ and $y$ in the cyclic group of order $6$. But we notice that for two cyclic groups with their order coprimes (for instance $36$ and $17$). This seems not to be a group homomorphism because we know that the only homomorphism between groups of coprime orders is the trivial homomorphism.

Could someone tell me why it is not a group homomorphism with out us Lagrange's theorem, but just use the definition of homomorphism. Which condition is this map does not satisfy?

Thank for any help!

• Did you try working out a small example yourself? Commented Aug 29, 2016 at 7:07
• I just find if it is a map defined as shown above there is something strange. I tried 36 an 17. And notice that x^51 should in the kernel if it is an homomorphism, since phi(x^51)=y^51=e. But if we use x^51=x^15 in the group of order 36, we have phi(x^51)=phi(x^15)=y^15 which is not the identity. I wonder why I cannot get the same answer. So I guess maybe something wrong here.
– Y.X.
Commented Aug 29, 2016 at 7:14

In general this will not even define a function between the groups, because the condition will require $\phi$ to take on different values at the same input.

(Notice, for instance, that if $G$ is cyclic of order 37 then there are many different ways of writing the identity: $$e = x^0 = x^{37} = x^{74} = x^{-37} = \cdots$$ Do these all give the same condition in general?)

• I think it is somehow relates to the example in my comment above.But could you tell me what is the point of difference between the 2 groups of coprime or non-coprime orders? Since in a cyclic group of order 12. We can also write the identity as different forms but phi IS a homomorphism between cyclic groups of order 12 and order 6.
– Y.X.
Commented Aug 29, 2016 at 7:24
• Well, in that case you can prove that all the different conditions actually turn out to be equivalent. But this isn't true in general. Commented Aug 29, 2016 at 7:25

$$\phi:C_{36}=\langle x\rangle\to C_{17}=\langle y\rangle\;,\;\;\phi(x^k)=y^k$$

For example we'd get

$$1=\phi(x^{36})=\phi(x^{17}x^{19})=\phi(x^{17})\phi(x^{19})=y^{17}y^{19}=1\cdot y^{19}=y^2\implies y^2=1$$

which of course is absurd.

Consider $Z_p$ and $Z_q$ where $p$ and $q$ are coprime.

As $Z_p$ and $Z_q$ are cyclic, we have that they each have some generator. Let's call them $x_p$ and $x_q$. We have that $x_p^p = e$, and $x_q^q=e$ (these are technically identities on different groups, but this won't be important).

Recall that any group homomorphism $\phi:G\to H$ has $\phi(e)=e$. Let $\phi:Z_p\to Z_q$ be a group homomorphism. Then, we have that: $$\phi(e)=e\implies \phi(x_p^p) = x_q^q\implies \phi(x_p)^p = x_q^q$$ We know that $\phi(x_p)\in Z_q$, and any element of $Z_q$ can be written as $x_q^m$ for some $m\in\mathbb Z/(q-1)\mathbb Z$. So, we have that $$x_q^q = x_q^{mp} = e$$

Here, we see that $mp = kq$ for some $k$. As $p$ and $q$ are coprime, it follows that $m = kq$. So, $\phi(x_p) = x_q^{kq} = e^k = e$, so $\phi$ is the trivial homomorphism.

• Could you tell me why p,q coprime and mp=kq implies m=kq?
– Y.X.
Commented Aug 29, 2016 at 7:40
• @YIMINGXU that was a mistake. It should be $m=k'q$ for some $k'$ strictly less than $k$. And this is true because $mp=kq$ means that $kq$ is divisible by $p$. $p$ and $q$ being coprime means $q$ isn't divisible by $p$, so $k$ must be. Commented Aug 29, 2016 at 7:51