Homomorphism between $Q_8$ and $S_4$

Question

Prove that there does not exist a homomorphism $\phi :Q_8 \rightarrow S_4$ such that $\phi (i) = (1 2 3 4), \phi (j) = (1 2 4 3) $.

Using homomorphism properties, we obtain $\phi(i) \phi(j) = \phi(ij) = \phi(k) = (1 2 3 4)(1 2 4 3) = (1 3 2)$, then $\phi(j) \phi(k) = \phi(jk) = \phi(i) = (1 2 4 3)(1 3 2) = (3 4)\neq (1 3 2)$, hence $\phi$ is not even a well-defined function. Is what I did right? I am not completely convinced. Thanks!

Answer 1

Your calculations are good and your proof is right, you could perhaps word it the following way to make the logic crystal clear:

• Let $\phi$ be some arbitrary function from $Q_8$ to $S_4$ that happens to take $i$ to $(1234)$ and $j$ to $(1243)$.

• Let's show that $\phi$ is not a homomorphism.

• If $\phi$ were a homomorphism, then $\phi(i) = \phi(jij) = \phi(j)\phi(i)\phi(j)$. But this implies $$ (1234) = (1243)(1234)(1243) $$ which is false: the right hand side fixes $1$ and the left hand side doesn't.

Answer 2

Your argument is correct. There is another, perhaps simpler argument: $$ (1234)^2=(13)(24),\qquad (1243)^2=(14)(23) $$ so $\phi(i^2)=(\phi(i))^2=(13)(24)$. On the other hand $\phi(j^2)=(14)(23)$. A contradiction, because $i^2=j^2=-1$.

Answer 3

In fact, it is not much harder to show that there is no homomorphism $\varphi:Q_8\to S_4$ such that $\varphi(i)=(1234)$. (In other words, we don't even need the hypothesis on $\varphi(j)$.)

Here is a sketch of the proof. Since $S_4$ does not contain a subgroup isomorphic to $Q_8$ (such a subgroup would be a Sylow $2$-subgroup, but those are dihedral), it follows that $\varphi$ cannot be injective. So the kernel of $\varphi$ is non-trivial. Now, $Q_8$ has a unique minimal normal subgroup, namely $\langle -1\rangle$, so this must be contained in the kernel. So $\varphi(i)^2=\varphi(i^2)=\varphi(-1)$ is the identity.