If $\phi$ is an isomorphism, $\phi(g)^n = 1 \iff g^n = 1$. Doesn't this hold for homomorphisms too?

Question

I need to prove that for an isomorphism $\phi$, the following is true:

$$\phi(g)^n = 1 \iff g^n = 1.$$

We know that

$$g^n = 1 \implies g\cdot g \cdots g = 1\implies \phi(g\cdot g \cdots g) = \phi(1) \implies \phi(g)\cdot\phi(g)\cdots\phi(g) = 1 \\\implies \phi(g)^n = 1.$$

The same argument can be applied to show that $\phi(g)^n = 1 \implies g^n = 1$.

But what can I say when $\phi$ is just an homomorphism? For me, everything is still valid. What's the problem?

Answer 1

If $\phi$ is a homomorphism, but not an isomorphism, then $g^n = 1 \implies \phi(g)^n = 1$ for exactly the reason you state, but the converse doesn't necessarily hold. That is, there could be $g \in G$ such that $\phi(g)^n = 1$ but $g^n \neq 1$. For example, consider $\phi : \mathbb{R}\setminus\{0\} \to \mathbb{Z}_2$ given by $x \mapsto \operatorname{sign}(x)$. Although $\phi(-2)^2 = 1$, $(-2)^2 \neq 1$.

If you write out the proof that $\phi(g)^n = 1 \implies g^n = 1$ for an isomorphism $\phi$, you will get to a point where $\phi(g^n) = 1$ from which you would like to conclude that $g^n = 1$. It is precisely here where the hypothesis that $\phi$ is an isomorphism is needed. As $\phi$ is an isomorphism, it has trivial kernel, so $g^n = 1$. If $\phi$ were just a homomorphism, you could not conclude that $\phi$ had trivial kernel; that is, $g^n$ may be some non-trivial element of the kernel (as in the example above).

Note that the only property of $\phi$ that was needed to prove $\phi(g)^n = 1 \implies g^n = 1$ is that $\phi$ has trivial kernel (i.e. $\phi$ is injective). While isomorphisms are injective, there are injective homomorphisms which are not isomorphisms (namely the non-surjective ones). So the logical equivalence $\phi(g)^n = 1 \iff g^n = 1$ holds for injective homomorphisms $\phi$, not just isomorphisms.

Answer 2

As the above comment already states, your proof holds for a homomorphism. But I think if you actually wrote out the other half you'd see why it must be an isomorphism to work. I'll do it for you.

$$1=\phi(g)^n=\phi(g)\cdot\phi(g)\dots\phi(g)=\phi(g\cdot g\dots g)=\phi(g^n)$$

Now if $\phi$ is an isomorphism then $\ker\phi=\{1\}$. And we see above $g^n\in \ker\phi$, so $g^n=1$. This doesn't always happen if $\phi$ is an arbitrary homomorphism.