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Prove that a group $G$ of order $280=2^3 \cdot 5 \cdot 7$ is not simple and has a subgroup of order 35

Using Sylow's theorem's it was easy to prove that $G$ is not simple.

I had a proof for the second statement, but I was unsure. The question "Proving if $|G|=280$, then $G$ is not simple" inspired me for a a different proof. Are both proofs correct?

Proof 1

"I'm pretty sure the following is correct"

Note $|\mathrm{Syl}_7(G)| \in \{1,8\}, |\mathrm{Syl}_5(G)| \in \{1,56\}$ and $|\mathrm{Syl}_2(G)| \in \{1,5,7,35\}$

If $|\mathrm{Syl}_7(G)| =1$ then the statement is easily proven. Choose a $P\in \mathrm{Syl}_7(G)$ which is also a normal subgroup and a $Q\in \mathrm{Syl}_5(G)$, then since $P\cap Q = 1$ the group $PQ\leq G$ and $|PQ|=|P|\cdot |Q| = 35$.

If $|\mathrm{Syl}_7(G)| =8$ then there are $8$ Sylow 7-subgroups of $G$. Now consider the action of $G$ on $\mathrm{Syl}_7(G)$ by conjugation. Choose a $P\in \mathrm{Syl}_7(G)$ and consider the orbit-stabilizer formula $$ |G| = |G_P|\cdot |P^G| \Rightarrow 280 = |G_P| \cdot 8 \Rightarrow |G_P| = 35 $$ Now since $G_P = \{g\in G: P^g= P\} = N_G(P) \leq G$ the group $N_G(P)$ is a subgroup of $G$ with the desired order.

Proof 2

"Less sure about this proof ..."

First notice that if $|\mathrm{Syl}_7(G)|=1$ or $|\mathrm{Syl}_5(G)| =1$ then choosing a $P$ in one and a $Q$ in the other Sylow $p$-subgroups once again creates a $PQ\leq G$ with order $35$.

Let nog $|\mathrm{Syl}_7(G)| \not =1, |\mathrm{Syl}_5(G)| \not = 1$, then because the group $G$ is not simple $|\mathrm{Syl}_2(G)|=1$. Let $P\in \mathrm{Syl}_2(G)$ then $P\trianglelefteq G$ with order 8.

Now consider the quotient group $G/P$ which has order 35. This group has through Sylow's theorems exactly one Sylow 5-subgroup, $A$ and one Sylow 7-subgroup $B$.

Notice how $A\trianglelefteq G/P$ and $|A|=5$ implies that $A$ is a cylic group with a generator $\langle Pa\rangle$ for a certain $a \in G$. Then $o(Pa) = 5$ which implies $a^5 \in P$. Because it is possible to choose 8 different $\tilde a\in G$ such that $Pa = P\tilde a$, choose now the $\tilde a$ such that $\tilde a^5=1$. Repeat for $B= \langle P\tilde b\rangle$ such that $\tilde b^7 = 1$. ($\color{red}{\text{? unsure if this is right}}$)

Now since $\langle \tilde a \rangle \times \langle \tilde b\rangle \cong \langle \tilde a\rangle \cdot \langle \tilde b\rangle$ it follows that $\langle \tilde a \rangle \cdot \langle \tilde b\rangle$ is a subgroup of $G$ with order 35.

Third proof (?)

Writing the last proof out, made me think of one shorter version of the last statement.

Since $5\mid |G|$ and $7\mid |G|$ then by Cauchy's theorem there is a $a, b\in G$ such that $o(a)=5, o(b)=7$. Now consider $\langle a\rangle \cdot \langle b\rangle$ which has trivial intersection and is a subgroup of $G$. (through isomorphism with a direct product)

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  • $\begingroup$ About the third proof, we would need that one of $\langle a\rangle$ and $\langle b\rangle$ is a normal subgroup or that they commute to conclude that their complex product is a subgroup. $\endgroup$ – Berci Dec 28 '17 at 22:40
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In your first proof second part you found the required group of order $35$ all right, but you fail to prove that $G$ is not simple, that is to find a non-trivial normal subgroup. Now you can reason as follows (write $n_p=\#Syl_p(G))$.

Assume that $G$ is simple. Then $n_5 \gt 1$ and $n_7 \gt 1$, whence $n_5=56$ and $n_7=8$. Since the Sylow $5$-subgroups intersect trivially pairwise and the same holds true for the Sylow $7$-subgroups, there are $(5-1)\cdot 56=224$ elements of order $5$ in $G$. Similary, there are $(7-1)\cdot 8=48$ elements of order $7$ in $G$. That leaves $280-224-48=8$ elements of $2$-power order. Since a Sylow subgroup of $G$ consists of $8$ elements, it follows that this one is unique hence normal, a contradiction.

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