**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,

D₅= < x, y| x⁵ = y2 = 1, yxy⁻¹ = x⁻¹ >

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,

D₆ = < x, y| x⁶ = y

^{2}= 1, yxy⁻¹ = x⁻¹ >

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,

D₇ = < x, y| x⁷ = y2 = 1, yxy⁻¹ = x⁻¹ >

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,

D₈ = < x, y| x⁸ = y2 = 1, yxy⁻¹ = x⁻¹ >

is a non-abelian group of order 16, we see that not every group of

order 16 is abelian.