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Let $K$ be the algebraic closure of $\mathbf{Z}/p\mathbf{Z}$,and let $w$ belongs to $Gal(K/(\mathbf{Z}/p\mathbf{Z}))$ be the Frobenius map sending $a$ to $a^p$.I have shown that $w$ has infinite order. I want an example of a map in $Gal(K/(\mathbf{Z}/p\mathbf{Z}))$ which does not belong to the cyclic subgroup generated by $w$.

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The Galois group is (isomorphic to) $\hat{\mathbb{Z}}$, so just pick your favorite element of $\hat{\mathbb{Z}}$ that is not contained in $\mathbb{Z}$, and that will give you the map you seek. If you aren't familiar with $\hat{\mathbb{Z}}$, then you should look into it! One keyword is "procyclic group". – Hurkyl Apr 19 '12 at 9:18
A related question. – Jyrki Lahtonen Apr 19 '12 at 9:22

For any integer $k=\prod_p p^{a(p,k)}$ ($p$'s are primes, $a(p,k)$ are their exponents in the factorization of $k$), let $0\leq g(k)\leq k-1$ be such that $p^{a(p,k)}\vert g(k)$ for every $p\neq 3$ and $g(k)\equiv1+3+3^2+3^3+\dots+3^{a(3,k)}$ mod $3^{a(3,k)}$. Such $g(k)$ exists and is unique by the Chinese remainder theorem.

If $F$ is the Frobenius automorphism, let $\phi\in Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)$ be defined by $\phi(t)=F^{g(k)}(t)$, where $k$ is such that $F^k(t)=t$ ($k$ depends on $t\in \overline{\mathbb{F}_p}$. This $\phi$ is not a power of $F$.

In fact, $Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)\cong\hat{\mathbb{Z}}$, where $\hat{\mathbb{Z}}$ is the profinite completion of the additive group $\mathbb{Z}$. By Chinese remainder theorem, $\hat{\mathbb{Z}}\cong\prod_p\mathbb{Z}_p$. You're asking for an element of $\hat{\mathbb{Z}}$ not in ${\mathbb{Z}}$, and I chose one ($0$ in $\mathbb{Z}_p$ for $p\neq 3$ and $1+3+3^2+3^3+\dots\in\mathbb{Z}_3$).

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Let $K_p\subset K= {\mathbb F_p}^{alg}$ be the the union of all $F_r=\mathbb F_{p^{l^r}} \; (r\geq 1)$ , where $l$ is a prime different from $2$ ( if $p\neq2$, you may even choose $l=p$).
The family of automorphisms $w^{n_r}\in Gal(F_r/\mathbb F_p), \; n_r =1+l+l^2+...+l^{r-1}$ is compatible and extends to an automorphism $W_0\in Gal(K_0/K)$.
The crucial point is that already $W_0$ is not a power of $w$, the Frobenius automorphism.
All you have to do now is to arbitrarily extend $W_0$ to some $W \in Gal(K/\mathbb F_p)$ by Steinitz and you have your required automorphism $W\notin w^{\mathbb Z}$ of $K$.

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a minor nitpick: you need to suppose $p\neq 2$ (otherwise $w^{n_r}=w^{-1}$) – user8268 Apr 19 '12 at 11:25
Dear @user, this is not nitpicking: this is doing me the great favour of correcting a genuine mistake. I have edited my answer in order to take the case $p=2$ into account. Many thanks for your careful reading and your comment. – Georges Elencwajg Apr 19 '12 at 13:49

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