The symmetric S7 acting on a set X with 5 elements

Question

I'm trying to prove that $S_{7}$ can't act transitively on a set with $5$ elements.

Now I know that the kernel of an action must be normal and all the normal subgroups of $S_{7}$ are $1$, $A_{7}$ and $S_{7}$ itself, so it must be one of them.

if its $A_{7}$ I can get contradiction

But I can't see why it can't be trivial, I tried to assume transitivity but I can't see where to use it to get contradiction

Answer 1

Recall that for $n\ge 5$, representations of $S_n$ other than $1$ and $\mathrm{sgn}$ have degree $\ge n-1$ (by e.g., the hook length formula.)

Thus, if $S_n$ acts on a set $X$ of size $k\le n-2$, the associated representation $\mathbb C[X]$ of $S_n$ must be a sum of $1$ and $\mathrm{sgn}$. Thus, the action of $A_n$ on $X$ must be trivial. But $S_n/A_n$ has size $2$, so if $k>2$ then no such nontrivial action can exist.

Answer 2

If $S_7$ acts transitively on a set with $5$ elements, a point-stabiliser would have index $5$ in $S_7$, but $S_7$ has no subgroup of index $5$.