# Carmichael lambda function proof

Let $$\lambda$$ be the Carmichael Lambda function, as defined on wikipedia. Let $$a,b\in\mathbb{Z}.$$ Prove the following

1. If $$a \mid b$$ then $$\lambda(a) \mid\lambda(b).$$
2. $$\lambda(\operatorname{lcm}(a,b)) = \operatorname{lcm}(\lambda(a), \lambda(b))$$.

I think the first statement can be shown using the unique factorization theorem and a case-by-case proof for the values of $$\lambda$$ on prime powers.

Wikipedia says that the second statement follows immediately from the recursive definition, but I don't think this is so; there's more to it than that. However, I'm not really sure how to prove the second statement by using the first definition more closely; if $$a$$ and $$b$$ are coprime then maybe using the unique factorization theorem will help but I'm not sure how to extend this result to when $$a$$ and $$b$$ are not coprime. I'd probably start by using this result for $$\frac{a}{\gcd(a,b)}$$ and $$\frac{b}{\gcd(a,b)}.$$

Hint  Like analogous proofs for Euler $$\phi$$ (totient) function, by multiplicativity we can reduce to the case of prime powers, where the proofs are straightforward. Alternatively, and more conceptually, we can use the universal properties of lcm and $$\lambda$$ (= universal order or expt). Let's prove $$(2)$$ in this manner. First, by $$(1)$$ and the lcm universal property we have, with $$[x,y] := {\rm lcm}(x,y)$$
$$a,b\mid [a,b]\,\Rightarrow\, \lambda(a),\lambda(b)\mid \lambda([a,b])\,\Rightarrow\, [\lambda(a),\lambda(b)]\mid \lambda([a,b])\qquad$$
For the reverse divisibility, if $$(x,[a,b])=1$$ then $$(x,a)=1=(x,b),\,$$ so $$\,x^{\lambda(a)}\equiv 1\pmod{\!a},\,$$ $$\,x^{\lambda(b)}\equiv 1\pmod{\!b},\,$$ so $$\,x^{[\lambda(a),\lambda(b)]}\equiv 1\pmod{[a,b]},\,$$ thus $$\,\lambda([a,b])\mid [\lambda(a),\lambda(b)],\,$$ by the universal analog of the Order Theorem, which shows that the minimal universal expt $$\,\lambda\,$$ divides every universal expt.