# Automorphism group of a group of order $p$

Prove that the automorphism group of a group of order $p$ is cyclic.

I have tried to solve this question for days but made no progress, can somebody help me with it?

a group of order $p$ is cyclic. So without loss of generality let the group be $\mathbb Z_p$

Then an automorphism can be uniquely determined by giving the image of $1$. How many choices are there? As many as elements of order $p$ in $\mathbb Z_p$ which happen to be $p-1$ how does composing to automorphisms work?

We just have to go where $1$ goes under the composition. Suppose $\phi$ send $1$ to $k$ and $\sigma$ send $1$ to $l$.

Then $\sigma \circ \phi$ send $1$ to $kl$. So the automorphism group of $\mathbb Z_p$ isisomorphic to $\mathbb Z_p^{*}$ (the multiplicative group mod $p$)

We shall now prove $\mathbb Z_p^{*}$ is cyclic of order $p-1$. The part about the order is because there are $p-1$ positive integers relatively prime to $p$ less than $p$

To see it is cyclic we use the fact the multiplicative group of any finite field is cyclic.

Proof of this statement: Let $F$ be a field of order $n+1$ and let $m$ be the least common multiple of the orders all of its non-zero elements. Then the polynomial $x^m-1=0$ has $n$ solutions(the order of the multiplicative group).

However it is a fact that a polynomial over a field cannot have more solutions than it's degree.

Therefore $n<m$ however by lagrange the order of all of the elements divides $n$. So we get $n=m$.

This means for every prime $p$ if $n=p^\alpha k$ with $p\nmid k$ there is an elment $g$ in the multiplicative group of $F$ with order a multiple of $p^\alpha$.

Therefore by the fundamental theorem of finitely generated abelian groups the $F\setminus\{0\}$ is a cyclic group.

I take as known that $$\operatorname{Aut}({\bf{Z}}/(p))\cong({\bf{Z}}/(p))^\times$$. Suppose there is $$K\le({\bf{Z}}/(p))^\times$$ such that $$K\cong C_q\times C_q$$, for some prime $$q$$. Therefore, there are $$q^2$$ elements $$x\in({\bf{Z}}/(p))^\times$$ such that $$x^q\equiv 1\pmod p$$: a contradiction, because this equation has at most $$q$$ solutions in $${\bf{Z}}/(p)$$. So, $$({\bf{Z}}/(p))^\times$$ does not contain any subgroup isomorphic to $$C_q\times C_q$$, for any prime $$q$$. For this characterization of the non-cyclic finite abelian groups, $$({\bf{Z}}/(p))^\times$$ must be cyclic.