# Describe a group homomorphism from $U_8$ to $S_4$

Im in an intro course to abstract algebra and we have been focusing completely on rings/the chinese remainder theorem and this question came up in the review and totally stumped me (we only have basic definitions of groups and subgroups and homomorphisms).

I think that $U_8$ is the group of units modulo 8, and $S_4$ is the permutation group of 4 letters. Ive figured out what $S_4$ looks like by examining certain sets of permutations but dont understand homomorphisms enough to be able to name the one in question. I do know that im looking for something of the form $f(ab) = f(a)f(b)$, but thats about it.

I was told a hint: that the units mod 8 were cosets which are relativley prime to 8, which i think would be $[1],[3],[5],[7]$ in mod 8, though im not really sure why this is the case. What I do notice is that each of these elements has an order of 2, which i think somehow should relate to the order of my permutations in $S_4$, but again, i'm not certain.

Any help is much appreciated, thanks.

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A unit mod $8$ is a congruence class mod $8$ which is invertible, i.e., a class $[a]$ such that there exists $[b]$ with $[a][b] = [1]$, or equivalently $ab +8k = 1$ for some integer $k$. Now any number dividing both $a$ and $8$ would also divide $ab+8k=1$, so this implies that $[a]$ being a unit implies $(a,8)=1$ (where the parentheses indicate the greatest common divisor.) On the other hand, one corollary of the Euclidean algorithm is that $(a,8)$ can always be written as a linear combination of $a$ and $8$, so in the case of relatively prime $a$ and $8$ there always exist such $b$ and $k$, and so $[a]$ is a unit.
If $f:U_8 \to S_4$ is a homomorphism, then the order of $\phi([a])$ always divides the order of $[a]$, so the image of $[1]$ has to be $()$ (the identity permutation), and the images of $[3]$, $[5]$, and $[7]$ have to have order $1$ or $2$. Obviously you also need that $f([3]) f([5]) = f([7])$ etc.