Prove all lie group homomorphisms $\phi:\mathbb{S}^1\to\mathbb{S}^1$ has the form $z\to z^n$ for some $n\in\mathbb{Z}$.
My idea - first of all, I cannot use the Lie algebra-lie group correspondence as we've yet to learn it. First I proved that every $1-1$ homomorphism of the circle group must be the identity or the inverse map. So it's left for me to prove that given such homomorphism $\phi$, it's kernel is the roots of unity of order $n$ for some $n\in\mathbb{N}$. My attempt:
Firstly, since $\ker(\phi)\subset\mathbb{S}^1$ is a closed subgroup, it is also a closed Lie subgroup. Assuming $\phi$ is not the trivial homomorphism, it is not both closed and open (as $\mathbb{S}^1$ is connected), and therefore must be $0$-dimensional submanifold (as $\mathbb{S}^1$ is one dimensional) - and consequently has the discrete topology. So $\ker(\phi)$ is a discrete subset of a compact set and hence finite.
Moreover, since every $z$ that is not root of unity has dense orbit in $\mathbb{S}^1$ and $\ker(\phi)$ is a closed proper subgroup, $\ker(\phi)$ can only contain roots of unity. Denote: $$N=max\{ord(z)\mid z\in\ker(\phi)\}$$ I want to claim that $\ker(\phi)$ is all roots of unity of order $N$. By choice of $N$, there exists $z\in\ker(\phi)$ with $ord(z)=N$, i.e: $$z=e^{\frac{2\pi i r}{N}}$$ with $r,N$ coprime. As $\ker(\phi)$ is a subgroup, $e^{\frac{2\pi i}{N}}\in\ker(\phi)$. This means that every root of unity satisfying $z^N=1$ is in $\ker(\phi)$. Assume there exists some $w\in\ker(\phi)$ with $z^m=1$ and $m\not\mid N$. Then $lcm(m,N)>N$, $zw\in\ker(\phi)$ and: $$(zw)^{lcm(m,N)}=1$$ This however does not lead me to a contradiction as there might be some $N\geq k\mid lcm(m,N)$ such that $ord(zw)=k$ - at least I wasn't able to prove there isn't.
How do I continute from here? I can't see the contradiction but if the statement in the question is true then there must be some way to reach contradiction. Any help would be appreciated.
I think you should just impose more. If $w\in \mathrm{ker}(\phi)$ such that $w^m=1$ then, passing to a power of $w$ if necessary, we can assume that $gcd(m,N)=1$, with $w^m=1$. The goal is to now show that this requires $m=1$. Indeed, as you say, $$(zw)^{mN}=1,$$ and I claim that no smaller power satisfies this (this uses $gcd(m,N)=1$ and $ord(z)=N$). This contradicts the maximality of $N$ unless $m=1$.