# Prove that if $|G| = 160$, $G$ is not simple.

I'm trying to prove this with Sylow's Theorem. I understand that the intersection between two Sylow-2 subgroups $$H$$ and $$K$$ cannot be of order $$16$$, since $$| H \cap K| = 16$$ implies $$H \cap K \lhd G$$. What I don't get, is how to proceed with the proof when $$|H \cap K| = 8$$. I did try to follow the proof at this link but I don't understand this part:

Suppose there were $$2$$ distinct Sylow $$2$$-Subgroups, and even if remaining three subgroups intersect in a group of order $$8$$, they are going to have too many elements.($$24 \times 3+7+31 \times 2$$=$$141$$ non-identity elements)

I think the calculation assumes that the three Sylow subgroups, say $$S_1, S_2, S_3$$, share the same intersection. But how can we be sure of this? Isn't it possible that $$|S_1 \cap S_2| = 8$$, $$|S_2 \cap S_3| = 8$$ but $$S_1 \cap S_2 \neq S_2 \cap S_3$$?

• I can see you might want as an exercise to do it this way, but @Sebastian Schoennenbeck 's comment on your linked question gives the easiest (and most useful in other contexts) way to solve the question. Commented Apr 16, 2021 at 6:58

Assume that $$G$$ is simple. Then it is easy to see that $$n_2(G)=5$$. So if $$P \in Syl_2(G)$$, then $$|G:N_G(P)|=5$$. $$G$$ acts by left multiplication on the left cosets of $$N_G(P)$$ and the kernel of this action is trivial, since $$G$$ is simple. It follows that $$G$$ can be embedded in $$A_5$$, violating $$|G|=160$$ not dividing $$60=|A_5|$$. See also here, Theorem (1.1) if you are unfamiliar with this argument.

• If $P$ is a Sylow 2-subgroup of $G$, then $|G:P|=5$ and $G$ acts on the left cosets of $P$ and then as you have. Commented Apr 16, 2021 at 8:43
• Yes correct, $N_G(P)=P$ in this case. Commented Apr 16, 2021 at 8:47

For the sake of future readers, I would like to post a more detailed solution.

By Theorem $$(3)$$ of the Sylow theorems, $$n_2 \in \{ 1,5\}.$$ If $$n_2=1$$, we are done because A unique Sylow $$p-$$subgroup is normal.

So, let's assume $$n_2=5$$ and $$Sly_2 (G)= \{ P_1, P_2, P_3, P_4, P_5\}.$$ Suppose $$G$$ acts on $$Sly_2 (G)$$ by conjugation; that is $$g.P_i=gP_ig^{-1}$$.

This action induces a homomorphism $$\phi: G \to S_5.$$ By Theorem $$(2)$$ of the Sylow theorems, $$\phi$$ cannot be a trivial homomorphism simply because its Sylow $$2-$$subgroups conjugate to each other. So, $$ker \phi \neq G.$$

On the other hand, $$160=|G| >|S_5|=120$$. So, $$\phi$$ is not injective. Thus, the kernel of $$\phi$$ is a proper normal subgroup of $$G$$ since The kernel of a homomorphism is a normal subgroup.

Note: In many problems an inequality such as $$|G| >|S_n|$$ (as above for $$n=5$$) doesn't hold so the kernel/injectivity trick doesn't work. In those cases, a possible approach is to show that the image of $$G$$ under $$\phi$$ must be a subgroup of $$A_n$$. For example, you can check out the solution by @Aryaman Maithaniat at this link.