$S_{10}$ has a subgroup of idex $4$

Question

Question:(True/False) $S_{10}$ has a subgroup of idex $4$.

Attempt I tried to use Lagrange's theorem $|S_{10}|=[S_{10}:H]|H|\Rightarrow |H|=\dfrac{10!}{4}$ but can't continue.

Any ideas?

Answer 1

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Following my comment and the other ones: Could it be $\;\ker\phi=A_{10}\;$ ? Observe that after answering this you obtain a straightforward contradiction by considering the size of $\;\ker\phi\;$ ...

Answer 2

Using the simplicity of $A_n$ for $n \ge 5$ to prove a result like this feels like using a sledgehammer to crack a nut, so here is a moderately elementary argument.

If $|G:H| = 4$ with $G=S_{10}$, then the action on cosets gives a homomorphism $\tau: G \to S_4$. Since $5$ does not divide $|S_4|$, all $5$-cycles lie in $\ker \tau$ and hence in $H$.

In particular, $h_1=(1,2,4,5,6)\in H$ and $h_2=(1,3,4,5,6) \in H$, and $h_1h_2^{-1} = (1,2,3)$ (or $h_2^{-1}h_1$ depending on how you compose permutations).

So $H$ contains all $3$-cycles, which generate $A_{10}$, and hence $A_{10}\le H$, contradicting $|G:H|=4$.