Suppose we have a homomorphism $f:G\to\Bbb Z$ for a finite group $G$. Prove that $f(g)=0$ for all $g$ in $G$

Question

Suppose we have a homomorphism from $G$, a finite group, to $\Bbb Z$, the set of integers, $f:G\to\Bbb Z$. To prove that $f(g)=0$ for all $g$ in $G$, what I did is defined an isomorphism - $h:G/\ker(f)\to{\rm im}(f).$ (First isomorphism theorem.)

Now, ${\rm im}(f)$ looks like any subset $m\Bbb Z$ of $\Bbb Z.$ Since $h$ is a 1-1 correspondence and maps the identities, we conclude that $m$ must be $0$ for this to hold, which means that the element going to identity is $\ker(f).$

But I am having trouble proving $\ker(f) = G$, I don't seem to notice what I'm missing. (If $\ker(f) = G$, then the problem is solved.)

Can someone please help me with this?

Answer 1

Hint: The order of $f(g)$ divides the order of $g$ for all $g\in G$. Note that $\Bbb Z$ is torsion-free.

Answer 2

Suppose for the sake of contradiction that there is some $g \in G$ for which $0 \neq f(g) \in \mathbb{Z}$. Next, consider the sequence:

$$f(g^1), f(g^2), f(g^3), \ldots$$

for which no two terms are the same: if $f(g) > 0$ then this is a monotonically increasing sequence, and if $f(g) < 0$ then this is a monotonically decreasing sequence. In any event, this sequence has no repeating elements.

However, $G$ is finite, which means $|G| = n$ for some positive integer $n < \infty$.

Then, $f(g^n) = f(e_G) = 0$, which contradicts the non-repetition of the sequence described above. Thus, our supposition was incorrect, and it must be that $f(g) = 0$ for all $g \in G$.

Answer 3

$\Bbb Z$ doesn't have any non-trivial finite subgroups: every non-trivial subgroup is infinite cyclic. But $G/\rm ker(f)\cong f(G)\le\Bbb Z$ is a subgroup. That's the homomorphic image of a group is a group.

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For a generalization of the first statement, any subgroup of a free abelian group on $n$ generators is free abelian on at most $n$ generators, hence infinite (or trivial). This is a theorem of Dedekind. You can find it in Lang's Algebra. Zorn's lemma is used.

Or, any subgroup of a free group is free: Nielsen-Schrier. This came later.

Answer 4

Suppose (towards a contradiction) that there exists $g\in G$ with $f(g) \neq 0$. Then, since $G$ is finite, $|G|=n<\infty$. Then the order of $g$ must divide $n$. In any case, $g^n = 1_G$, so $f(g^n) = f(1_G) = 0$. But this means that $f(g^n) = [f(g)]^n = 0$. Since $f(g)$ is an integer, we require that some nonzero integer $c$ exists such that $c^n=0$. But such an integer does not exist. Contradiction!