Homomorphisms $h:C_5 \to Aut(C_{31})$ (first course in group theory) I am asked to show explicitly all homomorphisms $h: C_q \to Aut(C_{31})$ for $q=5,3,2,29$
For $q=5$, $C_5 =\{1,y^2,...,y^4\}$ and $Aut(C_{31})=C_{30}=\{1,\alpha,...,\alpha^{29}\}$ so we know the order of $h(y)$ is 5. There are 4 elements of order 5 in $C_{30}$: $\alpha^6,\alpha^{12},\alpha^{18}$ and $\alpha^{24}$. But what is the strategy for identifying these explicitly in terms of automorphisms $\phi_a: x \mapsto x^a$? The only way I can think of is computing the orders of $\phi_a$ (all 30 of them...) and see which ones have order 5 but this can't be the best strategy(?!) I suppose I need to solve $a^5 $(mod 31) $=1$ but is there a way of doing this without going mental with the calculator? Finally, can I perhaps identify a generator $\phi_a$ of $Aut(C_{31})$ then the homomorphisms would be $\phi_a^6, \phi_a^{12},\phi_a^{18},\phi_a^{24}$ and what is a quick way of identifying $\phi_a$? Thanks. 
 A: By a theorem,  $\rm{Aut}(C_{31})$ is cyclic.
You will indeed need a primitive $\pmod{31}$, to describe these explicitly. There's no method for finding a primitive generally, so pick an element at random and check it.  Try $2$.  By a theorem,  it will be primitive precisely when $2^6\equiv 64\equiv 2\not\equiv 1$,  $2^{10}\equiv32\equiv 1$.  Oops, that means $2$ isn't primitive.   So try $3$.  $3^6\equiv 16\not\equiv 1$.  And $3^{10}\equiv 16\cdot 19\equiv 304\equiv-6\not \equiv 1$.  Finally $3^{15}\equiv -6\cdot 243\equiv 30\equiv-1$
Thus $3$ is primitive. So $\phi _3$ given by $\phi_3(1)=3$ generates $\rm{Aut}(C_{31})$.
Now, that means it has order $30$.  So the  $\varphi (5)=4$ elements of order $5$ are:  $3^6,3^{12},3^{18},3^{24}$.  These are $16,8,4,2$.
Now take $C_5$.  Any homomorphism is determined by $h(1)$.  So there are $5$.  They coincide with $h(1)=1,16,8,4,2$.
Use the same idea for $C_3,C_2$.  There's respectively $3$ and $2$ homomorphisms.
For $C_{29}$, things are particularly simple,  because $(29,30)=1$.  This means that there is only one homomorphism,  the trivial one.
A: Well here is one approach: pick a random element $b\neq 0 \pmod{31}$ and set $a\equiv b^6 \pmod{31}$. Unless you got unlucky and $b$ happened to be in the subgroup of order 6 (which only has probability $6/30=1/5$), then you'll have $a\neq 1$. Then
$$a^5 = (b^6)^5 = b^{30} = 1 \pmod{31}$$
so $\phi_{a}$ will be an automorphism of order 5, and the others will be $\phi_a^2, \phi_a^3$, and $\phi_a^4$.
A: A generator for $\text{Aut}(C_{31})$ is $\alpha: x \mapsto x^3$ ($3$ is a primitive element for $(\Bbb Z_{31})^{\times}$). So the $4$ elements of $\text{Aut}(C_{31})$ of order $5$ will be:
$\alpha^6: x \mapsto x^{16}$
$\alpha^{12}: x \mapsto x^8$
$\alpha^{18}: x \mapsto x^4$
$\alpha^{24}: x \mapsto x^2$
Don't forget to include the trivial homomorphisms $C_q \to \text{Aut}(C_{31})$ given by $h(y) = \text{id}_{C_{31}}$ for all $y \in C_q$.
