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

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.

• Use $\langle x\rangle$ for $\langle x\rangle$. Nov 20 at 21:22
• This has broken English. Try running the question through a translator. Nov 20 at 21:23
• I tried to correct the text to save the question but it didn't work out completely. Nov 21 at 5:16

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.