Group of Order 35 is Cyclic Group/Proof 2

Proof
Let $G$ be of order $35$.

We have that $35 = 5 \times 7$ where both $5$ and $7$ are prime.

Hence from the First Sylow Theorem, $G$ has:
 * at least one Sylow $5$-subgroup

and:
 * at least one Sylow $7$-subgroup

Let $n_5$ denote the number of Sylow $5$-subgroups of $G$.

Let $n_7$ denote the number of Sylow $7$-subgroups of $G$.

By the Fourth Sylow Theorem:
 * $n_5 \equiv 1 \pmod 5$

and from the Fifth Sylow Theorem:
 * $n_5 \divides 35$

where $\divides$ denotes divisibility.

It follows that $n_5 = 1$.

Let $P$ denote this unique Sylow $5$-subgroup.

By the Fourth Sylow Theorem:
 * $n_7 \equiv 1 \pmod 7$

and from the Fifth Sylow Theorem:
 * $n_7 \divides 35$

where $\divides$ denotes divisibility.

It follows that $n_7 = 1$.

Let $Q$ denote this unique Sylow $7$-subgroup.

From Sylow $p$-Subgroup is Unique iff Normal, $P$ and $Q$ are normal subgroups of $G$.

Consider $P \cap Q$.

By Intersection of Subgroups is Subgroup:
 * $P \cap Q$ is a subgroup of $P$
 * $P \cap Q$ is a subgroup of $Q$

and:
 * $P \cap Q$ is a subgroup of $G$

By Lagrange's Theorem:
 * $\order {P \cap Q} \divides \order P = 5$
 * $\order {P \cap Q} \divides \order Q = 7$

where:
 * $\order {P \cap Q}$ denotes the order of $P \cap Q$
 * $\divides$ denotes divisibility.

Thus:
 * $\order {P \cap Q} = 1$

and so:
 * $P \cap Q = \set e$

where $e$ is the identity of $G$.

From Prime Group is Cyclic, both $P$ and $Q$ are cyclic.

Let:
 * $P = \gen a$
 * $Q = \gen b$

where $\gen a$ denotes the subgroup of $G$ generated by $a$.

As $P$ is a normal subgroup of $G$:
 * $b^{-1} a^{-1} b \in P$

and so as $a \in P$, from :
 * $b^{-1} a^{-1} b a \in P$

As $Q$ is a normal subgroup of $G$:
 * $a^{-1} b a \in Q$

and so $b^{-1} \in Q$, from :
 * $b^{-1} a^{-1} b a \in Q$

Thus by definition of intersection:


 * $b^{-1} a^{-1} b a \in P \cap Q$

But $P \cap Q = \set e$.

So:
 * $b^{-1} a^{-1} b a = e$

and so from Product of Commuting Elements with Inverses
 * $a b = b a$

From Power of Product of Commutative Elements in Group:
 * $\paren {a b}^7 = a^7 b^7 = a^7 \ne e$

and:
 * $\paren {a b}^5 = a^5 b^5 = b^5 \ne e$

By Order of Element Divides Order of Finite Group:
 * $\order {a b} \divides 35$

where $\order {a b}$ denotes the order of $a b$.

Thus $\order {a b} \in \set {1, 5, 7, 35}$

As it has been established that $\order {a b}$ is not $1$, $5$ or $7$, it follows that:
 * $\order {a b} = 35$

Hence from Group whose Order equals Order of Element is Cyclic, $G$ is a cyclic group.