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Let G be a group of order $n$, where $n$ is a positive integer relatively prime to $\varphi(n)$. Show that G is cyclic.
You may only assume the Feit-Thompson theorem here and prove in the following way:
(1) $n$ is a product of odd prime numbers and squarefree.
(2) Then $G$ is solvable. Show that it has a cyclic quotient of prime order, that is there is a an epimorphism $G\to H$with $H$ cyclic of prime order. Let $N$ be the kernel. (Hint: using composition series)
(3)Show that $G\cong N \times H$ and then prove $G$ is abelian.
(4)Show that $G$ is cyclic.

I have proved (1) but get stuck at step 2. Is there any help? Thanks.

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If $G$ is solvable, then it has a solvable series, as $$G = G_0 \supseteq G_1 \supseteq \cdots \supseteq G_n = \{e\}$$ which can be refined to a solvable composition series. So we can assume $G_0, G_1, ... , G_n$ is a solvable composition series, i.e. $G_i \triangleleft G_{i+1}$ and $G_{i+1}/G_i$ is simple and abelian, i.e. cyclic of prime order. Thus $G/G_1$ is cyclic of prime order, and the canonical map $$G \rightarrow G/G_1$$ has kernel $G_1$. That's (2), and I'm not sure how to do part (3).

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