Find a monomorphism function $U_7\rightarrow S_5$

Question

Find a monomorphism $f$ function $U_7\rightarrow S_5$.

I know that:

$\phi(7)=6$

$|S_5|=120$

I tried to assign permutations but I'm not sure how:

$f(1)=id$

$f(2)=(1 2)$

$f(3)=(3 4)$

$f(4)=f(2)f(2)$

$f(5)=f(3)^-1$

$f(6)=f(2)f(3)$

Thank you

Answer 1

Any homomorphism would be determined by what it does to a generator of $U(7)$ (which happens to be cyclic). Choose an element of order six, say $(12)(345)$, to map the generator to.

The generators of $U(7)$ are $3$ and $5$.

Answer 2

Any cyclic subgroup of order $6$ of $S_5$ is a suitable image of $U(\Bbb Z/7\Bbb Z)\cong C_6=\{e,a,a^2,a^3,a^4,a^5\}$ under such an injective homomorphism. Such subgroups are all and only those of the form $\langle\sigma\rangle$, where $\sigma=(i_1,i_2)(i_3,i_4,i_5)$ with $\{i_1,\dots,i_5\}=\{1,\dots,5\}$ (as sets). As monomorphisms preserve elements order, we may have any of the following: $e\mapsto ()$, $a\mapsto \sigma$, $a^2\mapsto (i_3,i_5,i_4)$, $a^3\mapsto (i_1,i_2)$, $a^4\mapsto (i_3,i_4,i_5)$, $a^5\mapsto (i_1,i_2)(i_3,i_5,i_4)$.

Accordingly, there are $10$ such monomorphisms.

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Yet another approach based on group action. Such a monomorphism, $\varphi$, is equivalent to a faithful action of $C_6$ (the cyclic group of order $6$) on the set $X:=\{1,\dots,5\}$. The stabilizers are in particular subgroups of $C_6$, which are: $\{e\}$, $\{e,a^3\}\cong C_2$, $\{e,a^2,a^4\}\cong C_3$, $C_6$. Now:

1. none of the $5$ stabilizers can be $\{e\}$, because the respective orbit would have size $6$ $(>5$) (orbit equation: $5=6+\dots$);

2. none of the $5$ stabilizers can be $C_6$, because the intersection of all the stabilizers wouldn't be trivial and hence the action wouldn't be faithful (orbit equations: $5=1+1+1+1+1\Rightarrow \operatorname{ker}\varphi=C_6$, $5=1+1+1+2\Rightarrow \operatorname{ker}\varphi\cong C_3$, $5=1+1+3\Rightarrow \operatorname{ker}\varphi\cong C_2$, $5=1+2+2\Rightarrow \operatorname{ker}\varphi\cong C_3$).

Therefore, the only option left is to have $3$ stabilizers equal to $\{e,a^3\}$ and $2$ stabilizers equal to $\{e,a^2,a^4\}$ (orbit equation $5=2+3\Rightarrow \operatorname{ker}\varphi=\{e\}$). Choose any $3$ elements in $X$, say $i_3, i_4, i_5$; then, $\operatorname{Stab}(i_k)=\{e,a^3\}$, $k=3,4,5$, and hence $\varphi(a^3)$ is a permutation of order $2$ of $S_5$ which fixes $\{i_3,i_4,i_5\}$, namely $\varphi(a^3)$ is any $2$-cycle of $S_5$. Similarly, $\varphi(a^2)$ and $\varphi(a^4)$ are any distinct $3$-cycles of $S_5$ which fix $X\setminus\{i_3,i_4,i_5\}=\{i_1,i_2\}$.