# Proof: There can be no non-zero homomorphism from $(\Bbb Z_4,+)$ to $(\Bbb Z_5,+)$

Prove that there cannot be a non-zero homomorphism from the group $G_1=(\Bbb Z_4,+)$ to the group $G_2=(\Bbb {Z_5},+)$.

A homomorphism $f:G_1\to G_2$ should satisfy $f(x +_{4} y) = f(x) \ +_{5} \ f(y)$.

I know that the trivial homomorphism would be $f(t) = 0_5 \ \forall \ t \ \in G_1$, since in that case $f(x +_{4} y) = 0_5$ and also $f(x) \ +_{5} \ f(y) = 0_5 \ +_{5} 0_5 = 0_5$ for all $x,y\in G_1$.

But I'm not sure how to show that there can be no non-zero homomorphism. Isn't it possible that two elements of $G_1$ get mapped to the same element of $G_2$? Any ideas how to prove the general case?

P.S: In my notation $+_5$ refers to addition modulo $5$ and $+_4$ refers to addition modulo $4$. $0_5$ refers to the identity element in $G_2$.

• Hint: if you know $f(1)$, then you know the other values of $f$. In particular, $f(0) = f(1) + f(1) + f(1) + f(1) = ?$ – Connor Harris May 21 '18 at 15:26
• @ConnorHarris Yes, and then? I'm not sure I can see where you're going with this – user563280 May 21 '18 at 15:28
• You know that $f(0) = 0$ and $f(0) \equiv 4 f(1) \bmod 5$. What values of $f(1)$ allow both statements to hold? – Connor Harris May 21 '18 at 15:32
• @ConnorHarris Thanks, got it now! :) (after seeing Martin's answer) – user563280 May 21 '18 at 15:35

Any homomorphism $\phi$ is completely determined by the image of $1$.

That is, if $1 \mapsto a$, then $x \mapsto xa$. By Lagrange's theorem and this result " If $\vert g \vert$ is finite , then $\vert \phi(g) \vert$ divides $\vert g \vert$", we have $\vert a \vert$ divide both $4$ and $5$. So $\vert a \vert=1$ and hence $a=0$. So we have only the trivial map!

$f(5-_44)=f(5)-f(4)=f(1)=1=f(5)-f(0)=5f(1)=0$ since $4=0$ mod $4$, $f(4)=f(0)=0$ and $5=0$ mod $5$, $f(5)=5f(1)=0$.

First proof:

The image of $f$ is a subgroup of $\mathbb Z_5$; but this is simple, so either $f(\mathbb Z_4)=\mathbb Z_5$ (impossible, because the image has at most 4 elements), or $f(\mathbb Z_4)=\{0\}$.

Second proof:

Consider the element $f(1)$. We have $f(1)+f(1)+f(1)+f(1)=f(1+1+1+1)=f(0)=0$. So $f(1)$ is an element of order $4$ in $\mathbb Z_5$; but all elements in $\mathbb Z_5$ have order $5$, with the exception of zero. Thus $f(1)=0$, and it follows that $f=0$.

• Sorry, I couldn't understand your first proof. Are you saying that $\Bbb Z_5$ is a "simple group"? Could you prove it? Also how does the conclusion follow from there? – user563280 May 24 '18 at 23:39
• Yes, any group of prime order is simple. The proof is "simple": by Lagrange's Theorem, the order of a subgroup divides the order of the group. So when the order $n$ is prime, the only possible orders of subgroups are $1$ (the trivial subgroup) and $n$ (the full group). Just a little more thinking with this argument shows that if $G$ has prime order $p$ then $G\simeq\mathbb Z_p$. The conclusion follows as I said in the answer: $f(\mathbb Z_4)$ is a subgroup of $\mathbb Z_5$ with 4 elements or less; the only possibility is 1. – Martin Argerami May 25 '18 at 0:21

I’m not sure what you are looking for but if you’re allowed LaGrange’s Theorem then the following should work:

Suppose that $f : G_1 \rightarrow G_2$ is a group homomorphism. Then $f(G_1)$ is a subgroup of $G_2$ and so we may apply LaGrange’s Theorem to say that the order of $f(G_1)$ must divide the order of $G_2$. Since $|G_2| = 5$ and $|G_1| = 4$, it follows that $|f(G_1)| = 1$ which is to say that $f$ must be the trivial homomorphism.

• Why should $f(G_1)$ be a subgroup of $G_2$? Could you point out the proof (or give me a link to the proof) to me? – user563280 May 21 '18 at 15:35
• @Blue the range (or image, whichever you prefer) of a homomorphism is a subgroup of the group being mapped to. Textbooks tend to leave it as an exercise since it’s just a matter of verifying that the conditions of the subgroup test hold. – user328442 May 21 '18 at 15:37