Characteristic Subgroups in Finite Cyclic Groups Suppose $E = <e>$ is a cylic subgroup of $C = <c>$ such that $e=c^j$. How can I show that for any automorphism $F: C \rightarrow C$ that $e \in f(E)$ i.e that $E \subset F(E)$?
 A: $\newcommand{\Z}{\mathbb{Z}}$ Ok, I'm going to pick some different notation because I want to use $e$ for the identity. Let $G$ be a cyclic group of order $n$ generated by $g$, so it is canonically isomorphic to $\Z_n$ under the isomorphism that sends $g$ to $1$ call this isomorphism $f$. Then let $H$ be the cyclic subgroup of $G$ of order $j$ generated by $h$. We have that $H$ sits as a subgroup of $\Z_n$ under $f$ where $f(h) = m$. Every automorphism of $\Z_n$ is given by the map that sends 1 to an element of $A = \{ a : 0 < a < n, \gcd(a,n)=1 \}$. If $a \in A$ then $h(k) = ak$ is an automorphism of $\Z_n$, we need to prove that the set $\{ab : b \in f(H) \}$ contains $f(H)$. Once you've reduced this to the $\Z_n$ case, the rest of the proof is using Lagranges theorem and a coprimality condition on $b$... Do you see it?
A: Well, one way to get at this would be to recall two facts:
1) $E$ is the unique subgroup of $C$ of order $|E|$ (cyclic groups have exactly one subgroup of order $d$ for each divisor $d$ of $|C|$).
2) Automorphisms preserve the order of an element.  That is, if $f$ is an automorphism of $C$, then the order of $f(e)$ is equal to the order of $e$.
Fact 2 implies that the order of $f(e)$ is equal to the order of $e$, and therefore $f(e)$ generates a cyclic subgroup of order $|E|$.  In other words, $|<f(e)>| = |<e>| = |E|$.  Since there is only one subgroup of order $|E|$ (which is of course $E$), we can conclude that $f(E)$ must be equal to $E$.
