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.


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.

Now the question is what exactly you are trying to find, just one homomorphism (which is easy, there is always the trivial one mapping everything to the identity), or all of them (which is not quite as easy but doable with the information here and some trial and error.)

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