Group of Order 35 is Cyclic Group/Proof 2
Theorem
Let $G$ be a group whose order is $35$.
Then $G$ is cyclic.
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:
and:
- $P \cap Q$ is a subgroup of $G$
- $\order {P \cap Q} \divides \order P = 5$
- $\order {P \cap Q} \divides \order Q = 7$
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 Group Axiom $\text G 0$: Closure:
- $b^{-1} a^{-1} b a \in P$
As $Q$ is a normal subgroup of $G$:
- $a^{-1} b a \in Q$
and so $b^{-1} \in Q$, from Group Axiom $\text G 0$: Closure:
- $b^{-1} a^{-1} b a \in Q$
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.
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: Exercise $1$