Let $G = S_4$ and let $a=(12), b=(234)$. Then there is a unique homomorphism from $G$ to $H = S_3$ that satisfies $f(a)=(12)$ and $f(b) = (123)$. Show a diagrammatic representation of it.
I have no clue on how to do this problem. Any help on where to start?
Note that $S_4$ has only one subgroup of order $12$, the alternating group $A_4$. You can find the proof of this fairly easily.
Let $K$ denote the subgroup of $S_4$ generated by $(12),(234)$. Then $K$ contains an element of order three (namely, $(234)$) and also an element of order four (namely $(12)(234) = (1234)$), so that $|K|$ is divisible by $12$. Because $|K|$ must divide $|S_4| = 4! = 24$, the only possibilities for $|K|$ are $12,24$.
If $|K|=12$ then necessarily $K = A_4$. But this can't be the case since $K$ contains elements of even and odd sign. Thus $|K| = 24$ and $K = S_4$. That is to say, $(12),(234)$ generate $S_4$. Any homomorphism is uniquely determined by its behavior on a generating set for its domain.
