Show that $S_5$ does not have a quotient group isomorphic to $S_4$

Question

Show that $S_5$ does not have a quotient group isomorphic to $S_4$.

If we to assume that $H$ is such a group, than $H$ must be normal in $S_5$ and $|H|=|S_5|/|S_4|=5$. So $H$ must be isomorphic to $\mathbb{Z}/5\Bbb Z$.

That's as far as my logic goes. I couldn't arrive at a contradiction.

Any ideas?

Answer 1

The possible candidates for such an $H$ are the subgroups of $S_5$ that are cyclic of order 5. All elements of $S_5$ of order 5 are given by $5$-cycles. However, the subgroup generated by a 5-cycle is not normal, so no $H$ can exist, as desired.

Answer 2

Let H be normal subgroup of $S_5$ such that $S_5/H$ is isomorphic to $S_4$.

Then $H=<\sigma>$ for some 5 cycle $\sigma$.

Note that $\tau(a_1\ a_2\ a_3\ ....\ a_k)\tau^{-1} = (\tau(a_1)\ \tau(a_2) ... \ \tau(a_k))$.

It is very useful find $\tau \in S_5$ s.t $\tau H \tau^{-1}$ is not $H$

Answer 3

The kernel of a surjective homomorphism from $S_5$ to $S_4$ would have order $|S_5|/|S_4|=5.$ This is impossible because: $S_5$ has $1+4!=25$ elements of order $1$ or $5$; the image of each of those $25$ elements must have order $1$ or $5$ in $S_4$; but $S_4$ has no elements of order $5,$ so those $25$ elements must all belong to the kernel of the homomorphism.