If $l \equiv 1$ mod $p$, then $l^{p^{e-1}}\equiv 1$ mod $p^e$.

Question

I am in the following hypothesis: assume that $A$ is an abelian p-group of finite exponent and suppose that $\alpha$ is a power automorphism. It can be shown that there exists a positive integer $l$ such that $\alpha(a)=a^l$ for all $a \in A$. Now i want to show that if $\alpha$ is nontrivial and has order prime to $p$, then $\alpha$ is fixed-point-free. Equivalently i want to show the contrapositive. Suppose that $\alpha \neq 1$ is not fixed-point-free: then $l \equiv 1$ mod p and so $l^{p^{e-1}}\equiv 1$ mod $p^e$, where $p^e$ is the exponent of $A$. I don't understand the last congruence... Probably it is stupid but i can't see it. Thank you to everybody.

Answer 1

You can prove it by induction:

Suppose that, for some $e\ge 1$, we have $$ l^{\mkern1mu p^{\scriptstyle e-1}}\equiv 1 \mod p^e\quad\text{i.e. }\exists\,m\in \mathbf Z,\:l^{\mkern1mu p^{\scriptstyle e-1}}=1+m\, p^e .$$ Then $$l^{\mkern1mu p^{\scriptstyle e}}=\Bigl(l^{\mkern1mu p^{\scriptstyle e-1}}\Bigr)^p=\bigl(1+m\, p^e\bigr)^p=1+\sum_{k=1}^p\binom pk m^kp^{k\mkern 2mue}$$ Now observe that, if $k\ge 2$, $\,ke\ge e+1$. On the other hand, the first term in the above sum is $\,pmp^e=mp^{e+1}$, so all terms in the sum are multiples of $p^{e+1}$. This proves that $$l^{\mkern1mu p^{\scriptstyle e}}\equiv 1\mod p^{e+1}$$ and ends the inductive step.