The group $S_7$ has no subgroups of order $840$.

Question

good afternoon everyone. I was studying group theory and came across the following exercise:

Let $x=(3,4,8,6)(5,7)$, $y=(2,1)(8,6)(5,7)(3,4)$, and $G=\langle x,y\rangle$.

• Denote by $A,B,C$ and $D$ the four right cosets of the subgroup $H=\langle y\rangle$ in $G$ and compute an element of each one.
• Let $\varphi$ be the action of $G$ by right multiplication on $\{A,B,C,D\}$. State which permutations are $\varphi (x)$ and $\varphi (y)$.
• Prove that $G$ is isomorphic to $\langle\varphi(x),\varphi(y)\rangle$.
• Finally, argue that $S_7$ has no subgroups of order $840$ (hint: consider the action on cosets and consider the normal subgroups of $S_7$).

For the first three parts, I proceeded as follows:

I have that $\Omega=\{A=H, B=\{x,yx\}, C=\{x^2,yx^2\}, D=\{x^3,yx^3\}\}$. The action of $G$ on this set is:

$\varphi: G \rightarrow S_{\Omega}$, given by $g \rightarrow Hg$

Then $\varphi (x) = Hx = B$ and $\varphi (x) = Hy = H = A$.

We can observe that since $\langle\varphi(x),\varphi(y)\rangle$ generates $S_\Omega$ and has the same elements as $G$, we can consider the identity isomorphism to see that they are isomorphic. $G=\{e,x,x^2,x^3,y,yx,yx^2,yx^3\}$

Is this reasoning correct, or is something missing? On the other hand, I do not know how to proceed to prove the last part.

Answer 1

By contradiction, let $H\le S_7$ have order $840$. $S_7$ acts by left multiplication on the left quotient set $S_7/H$, of size $7!/840=6$. The kernel of this action is the intersection of all the conjugates of $H$ in $S_7$, and must have order $1$ or $7!/2$ or $7!$, because the only normal subgroups of $S_7$ are $\{()\}$, $A_7$ and $S_7$ itself. The two latter are ruled out because they are too big to turn out as intersection of subgroups of order $840$. The former is ruled out because it would give rise to an embedding $S_7\hookrightarrow S_6$, which is impossible for cardinality reasons.