SOME IMPORTANT THEOREMS ON CYCLIC GROUP

Theorem: Let G be a finite group of order n. If G contains an element of order n, then G must be cyclic.
Proof: Let a ∈ G such that O(a) = n.
Let H = {a^r : r ∈ 1 } be a subgroup of G.
But O(a) = n
⇒ H = {e, a, a², ... , a^n-1} = <a>
i.e., H is a cyclic subgroup of G generated by a.
Also O(H) = O(G)
⇒ G = H = <a>.
i.e., G is a cyclic group.

Theorem: Every cyclic group is abelian.
Proof: Consider a cyclic group G  generated by a.
i.e.,  G = <a>.
Let x, y ∈ G be arbitrary element.
∴ x = aⁿ and y = a^m for some integers n and m.
Then xy = aⁿa^m = a^n+m =a^m+n = a^maⁿ = yx.
∴ G is an abelian group.

Theorem: Prove that a subgroup of a cyclic group is cyclic.
Proof: Let G = <a> be a cyclic group generated by a.
Let H be a subgroup of G.
If H = G, then H = <a> is a cyclic group generated by a.
If H = {e}, then H = <e> is a cyclic group generated by e.
So, let H ≠ G, {e} i.e.,  h is a proper subgroup of G.
∴ ∃ an element x ∈ H such that x ≠ e.
Now x ∈ H
⇒ x ∈ G.
⇒ x = a^r for some non-zero integer r
∴ x ∈ H
⇒ x^-1 ∈ H, since H is a subgroup of G.
⇒ a^-r ∈ H
∴ a^r , a^-r ∈ H.
Since r ≠ 0, therefore atleast one of r, -r is a positive integer.
So, positive integral powers of a belong  to H.
Let q be the least positive integer such that a^q ∈ H.
We shall prove that H is a cyclic group generated by a^q.
Let x ∈ H be arbitrary element
⇒ x ∈ G                                                                           [∵ H ⊆ G]
⇒ x = aⁿ  for some integer n.                                                 ... (1)
By division algorithm, ∃ integers s and r such that
           n = sq + r     where 0 ≤ r ≤ q
⇒      aⁿ = a^{sq + r}

⇒      aⁿ = a^sq a^r
⇒     a^{n- sq} = a^r .                                                                       ... (2)
Since a^q ∈ H and s is an integer, so a^sq∈ H.
Also aⁿ ∈ H
∴ aⁿ (a^sq)^-1 ∈ H
⇒ aⁿ a^{- sq} ∈ H
⇒ a^n-sq ∈ H
⇒ a^r ∈ H,                                                                           From (2)
∴ a^r ∈ H    0 ≤ r ≤ q-1.                                                           ... (3)
But q is the least positive integer such that a^q ∈ H.
∴ r = 0,                                                              [From (2) and (3)]
∴ From (1), (2) and (3) we get
     x = aⁿ = a^{sq + r} = a^{sq + 0} = (a^q)^s
This is true for all x ∈ H.
∴ Each element of H is an integral power of a^q.
So, H is cyclic group generated by a^q.
Hence subgroup of a cyclic group is cyclic.

Theorem: Every group of prime order is cyclic.
Proof: Let G be a group of order p, a positive prime.
Since p ≥ 2
∴ G has atleast two elements.
Consider a ≠ e ∈ G and let H = <a> be the cyclic subgroup of G.
Therefore O(H) = O(a) > 1.
By Lagrange's theorem O(H) | O(G)
i.e.,  O(H) | p.
But p is prime.
Therefore O(H) = p = O(G).
Hence G must be cyclic group.