Using the number of Sylow 2-subgoups of $G$.

Question

I want to show that a group of $G$ of order $56$ is not simple using the number of Sylow $2$-subgroups of $G$ with $n_{2}=7$ and considering two Sylow $2$-subgroup $P_{1},P_{2}$ of $G$.

(I know the way counting certain elements of order $7$ assuming $n_{7}=8$. Just to try the other ways in this problem.)

I guess the relation between $P_{1}\cap P_{2}$ and $N_{G}(P_{1}\cap P_{2})$ giving a way to show this problem.

But, i can't find any clue.

Anyone can give some advice? Thank you!

Answer 1

Assume that $n_2=7$, and let $P_1,P_2$ be two Sylow $2$-subgroups. Then $|P_1\cap P_2|\in\{1,2,4\}$. Since $|P_1P_2|=\frac{|P_1|\cdot|P_2|}{|P_1\cap P_2|}=\frac{64}{|P_1\cap P_2|}\leqslant 56$, we see that $|P_1\cap P_2|\neq 1$. Hence $|P_1\cap P_2|\in\{2,4\}$, but $|P_1P_2|\in\{16,32\}$ follow as well.

We can further conclude $G=\langle P_1,P_2\rangle$. Indeed, $8\mid |\langle P_1,P_2\rangle|$ since $P_1\leqslant \langle P_1,P_2\rangle$, $|\langle P_1,P_2\rangle|\mid 56$ obviously, but also $P_1P_2\subseteq\langle P_1,P_2\rangle$, and $P_1P_2$ has at least $16$ elements. The conclusion is that $\langle P_1,P_2\rangle=G$ is the only option.

Case $|P_1\cap P_2|=4$. Then $[P_i:P_1\cap P_2]=2$, so $P_1\cap P_2\lhd P_i$, for $i=1,2$. Therefore $P_1,P_2\leqslant N_G(P_1\cap P_2)$, so $G=\langle P_1,P_2\rangle\leqslant N_G(P_1\cap P_2)$. Thus $P_1\cap P_2\lhd G$ and $G$ is not simple.

Case $|P_1\cap P_2|=2$. There are subgroups $P_1',P_2'$ of order $4$, such that $P_1\cap P_2\leqslant P_i'\leqslant P_i$ for $i=1,2$. Of course, $P_1\cap P_2\lhd P_i'$ as these are group extensions of index $2$, so $P_1',P_2'\leqslant N_G(P_1\cap P_2)$, and consequently $\langle P_1',P_2'\rangle\leqslant N_G(P_1\cap P_2)$.

Also $P_1\cap P_2=P_1'\cap P_2'$ because otherwise we would get $|P_1\cap P_2|>2$. So $|P_1'P_2'|=\frac{|P_1'|\cdot |P_2'|}{|P_1'\cap P_2'|}= 8$. By reasoning as above we conclude $4\mid|\langle P_1',P_2'\rangle|\mid 56$ and $|\langle P_1',P_2'\rangle|\geqslant 8$, so the options are $|\langle P_1',P_2'\rangle|\in\{8,28,56\}$.

Subcase $|\langle P_1',P_2'\rangle|\in\{28,56\}$. Then $\langle P_1',P_2'\rangle$, and hence $N_G(P_1\cap P_2)$, contains an element $a$ of order $7$. Since $P_1\cap P_2=\{e,b\}$, we get $b^a=b$, so $a^b=a$ as well, which means that Sylow $7$-subgroup $\langle a\rangle $ has a normalizing element $b$ of order $2$. By Sylow theorem we get $n_7=1$ and $\langle a\rangle\lhd G$.

Subcase $|\langle P_1',P_2'\rangle|=8$. Then $P_3=\langle P_1',P_2'\rangle$ is a Sylow $2$-subgroup different from both $P_1$ and $P_2$. Then $P_1\cap P_3=P_1'$ is of order $4$, so we are done by the first case.

Answer 2

Suppose the number of Sylow $7$ subgroup is $1$. then $G$ is not simple. Else number of Sylow 7 subgroup must be 8. Sylow 7 subgroups are cyclic of prime order hence can intersect only at the identity element. Then there are $6*8=48$ elements of order 7. The rest must be elements in the Sylow 2 subgroup, which means that there is only one Sylow 2 subgroup hence showing again that the group is not simple.