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Let $G$ be a group such that for all $a$, $b$ $\in$ G, there exists $f$$\in$ Aut(G) satisfy $b$=$f(a)$.

i) Prove that if there is nontrivial élément in $G$ of finite order, then there exist prime number $p$ such that every nontrivial element in $G$ has order $p$

ii) Prove that if G is finite,then G is abelian p-group

I guess that we prove i) using cauchy theorem, but for ii) i have no idea how to prove it. I nead help to prove ii) in detail by giving some hints

Thank you.

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$(i)$ Elements $a,b$ with $b=f(a)$ have the same order. Hence by hypothesis all elements except the identity have the same order. Let $p$ be a prime divisor of $|G|$. By Cauchy's theorem there exists an element of order $p$. Hence all non-trivial elements have order $p$.

$(ii)$ Since $G$ is a non-trivial $p$-group by $(i)$, it has a non-trivial center. Elements $a,b$ with $b=f(a)$ are either both in the center or both not in the center. Hence by hypothesis every non-trivial element is in the center. Hence $G$ is abelian, i.e., an abelian $p$-group.

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Let $x$ be an element of finite order $m$, and $p$ a prime which divides $m$, write $y=x^{m/p}$ the order of $y$ is $p$. Let $z\in G$, $z=f(y)$, $z^p=f(y^p)=1$.

Suppose that $G$ is finite, if $x$ has order $m$, if $p,q$ are prime which divide $m$, the order of $y=x^{m/p}$ is $p$, the order of $z=x^{m/q}$ is $q$. Write $y=f(z)$, $y^q =f(z^q) =1$. This implies $q$ divides $p$, thus $q=p$ since $p$ and $q$ are prime. The part 1 implies that G is a p-group

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