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Every Cyclic Group is Abelian But The Converse is not necessarily TRUE

MATHS TRICK-20
This is a trick On How To Check any abelian group whether it is Cyclic Group or not.

Trick: Every cyclic group is Abelian.
Solution: Suppose that G is a cyclic group that is generated by the element g.
Let x and y be arbitrary elements of G. we must show that xy = yx.
Since G is generated by g, there must exist integersr and s such that
x = g^r , y = g^s. But then xy = g^r.g^s= g^(r+s)= g^s.g^r = yx

Important Notes:-
<i> The order of cyclic group is same as the order of its generators.
<ii> Every group of Composite Order is not Cyclic.

Illustrative Example: For each n, 9 ≤ n ≤ 16, answer the following questions:
(a): Is every group of order n cyclic?
(b): Is every group of order n abelian?
(c): Is every abelian group of order n cyclic?

Answer: Consider n = 9. Then, since 3 is not relatively prime
to itself, Z/3 × Z/3 is not cyclic, so we see that not every group of
order 9 is cyclic and not every abelian group of order 9 is cyclic.
Since these are the only abelian groups of order 9 and we know that
groups of order p2 for a prime p are abelian, this comprises the entire set of groups of order 9, so we can say that every group of order 9 is abelian.

Consider n = 10. Then Z/10  Z/2 × Z/5 since (2, 5) = 1 and,
since these are the only abelian groups, we see that every abelian
group of order 10 is cyclic. Now,
 D= < x, y| x = y2 = 1, yxy¹ = x¹ >
is of order 10 but isn’t abelian, so we see that every group of order
10 is not abelian and, thus, not all groups of order 10 are cyclic.

Consider n = 11. Then every subgroup of a group of order 11
must be of order 1 or 11, so each element except the identity is of
order 11, so the only group of order 11 is Z/11.

Consider n = 12. Then, since (3, 4) = 1, Z/12 ≅ Z/3 × Z/4.
However, since Z/4 is not isomorphic to Z/2 × Z/2, we see that
Z/12 is not isomorphic to Z/3 × Z/2 × Z/2, which isn’t cyclic, so we
know that not every group of order 12 is cyclic and not every abelian
group of order 12 is cyclic. Furthermore,
D = < x, y| x = y2 = 1, yxy¹ = x¹ >
is of order 12 and non-abelian, so we see that not every group of
order 12 is abelian.

Consider n = 13. By the same reasoning given in the case where
n = 11, we see that Z/13 is the only group of order 13, so the answer
to all the questions is “yes”.

Consider n = 14. Since (2, 7) = 1, Z/14  Z/2 × Z/7 and, since
these are the only abelian groups of order 14, we see that every
abelian group of order 14 is cyclic. Now,
D = < x, y| x = y2 = 1, yxy¹ = x¹ >
is a non-abelian group of order 14, so not every group of order 14 is
abelian or cyclic.

Consider n = 15. Since (3, 5) = 1, Z/15  Z/3 × Z/5 and,
since these are the only abelian groups of order 15, we see that
every abelian group of order 15 is cyclic. Furthermore, since 3 does
not divide (5 − 1) = 4, we can conclude that any group of order 15 is abelian and, therefore,
cyclic.

Consider n = 16. Then Z/16  Z/4 × Z/4, so not every (abelian)
group of order 16 is cyclic. Furthermore,
D = < x, y| x = y2 = 1, yxy¹ = x¹ >
is a non-abelian group of order 16, we see that not every group of
order 16 is abelian.