Prove that Euler phi function is multiplicative by a given theorem

Question

I had proven a theorem which states that
If $G=\langle a\rangle$ has order $rs$ , where $(r,s)=1$. Then there are unique $b,c\in G$ with $b$ of order $r$, $c$ of order $s$ and $a=bc$.

There is another theorem which states that
If $G=\langle a\rangle$ is cyclic of order $n$, then $a^k$ is also a generator of $G$ if and only if $(k,n)=1$.

Although I know that there are many ways to prove this statement for example by using Chinese Remainder Theorem, but I am required to use the first theorem to establish the proof. So I wonder what is the relationship between the theorem and the statement. Can I say that $b$ and $c$ are also generator of $G$?

Answer 1

You can't conclude that $b$ is the generator of $G$ because $b$ has order $r$ which is less than order of G which is $rs$. Order of generator must be equal to order of a group $G.$ Similar argument for $c$.

Answer 2

Hint: by Bézout's Theorem one can find $k, l \in \mathbb{Z}$ with $kr+ls=1$. Note that gcd$(k,s)=1=$gcd$(l,r)$. Hence $a=a^1=a^{kr+ls}=(a^k)^r \cdot (a^l)^s$. Put $b=a^{ls}$ and $c=a^{kr}$.