# Thread: Groups of order 27

1. ## Groups of order 27

Hello,

I am given that G is a finite abelian group that contains
8 elements of order three
18 elements of order nine
and an identity

And am told to describe of possibilities for G, by giving explicit decomposition into cyclic groups up to an isomorphism

So it seems like there are three possibilities for G

$\mathbb{Z}_{27}$
$\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$
$\mathbb{Z}_9 \times \mathbb{Z}_3$ (is this isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_9$?)

Now G cannot be $\mathbb{Z}_{27}$, for G has no element of order 27.

I also think G cannot be $\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$, since no element of that group has order nine. I pretty sure every element of this group must have order less than or equal to 3, since if $(a,b,c) \in$ $\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$, then $3*(a,b,c) = (0,0,0)$

So I am left with $\mathbb{Z}_9 \times \mathbb{Z}_3$. Is there a quicker way to test the elements of this group besides considering the order of all 27 elements?

i am pretty sure the element (1,a) \in $\mathbb{Z}_9 \times \mathbb{Z}_3$, would have order 9, but I dont think there's 18 of those.

2. ## Re: Groups of order 27

It is easy to see that G is $\cong$ to $Z_3 \oplus Z_9$ for the order of an element of this group if the lcm(a,b), where $a \in Z_3, b \in Z_9$ $Z_9$ has elements of order 9,3,1 (order of elements divides order of group, since cyclic these are the only 3 possible). So since $Z_9$ is cyclic, if has $\phi(9) = 6$ generators, or elements of order 9. So basically 6 * 3, elements of $Z_3 \oplus Z_9$ would be of the form (a,9), lcm(a,9) = 9. so 6*3 = 18 elements of order 9. 1 element of order 1, so 27-19, elements of order 3 (the only other possible order). or 8 elements of order 3.

3. ## Re: Groups of order 27

yes it has to be Z9xZ3 (which is isomorphic to Z3xZ9: it is easy to check that (a,b)-->(b,a) is an isomorphism).

in general, for an abelian p-group (a finite group whose order is a power of a prime p), which p-group we have can be ascertained by examining how many elements of order p we have (this is particularly useful for p = 2).

Z27, being cyclic, has just ONE subgroup of order 3, and therefore just TWO elements of order 3.

Z3xZ3xZ3 has 26 elements of order 3.

clearly, if (a,b) in Z9xZ3 has order 3, either a has order 3, or b has order 3, or both.

ONLY a has order 3:

(3,0)
(6,0)

ONLY b has order 3:

(0,1)
(0,2)

a and b BOTH have order 3:

(3,1)
(3,2)
(6,1)
(6,2)

there are your 8 elements of order 3. this is all of them.

similarly reasoning applies to elements of order 9: a must have order 9, b might be of order 1, or 3 (in other words b can be anything). this gives us φ(9) = 6 choices for a, the 6 generators of Z9:

1,2,4,5,7 and 8. so explicitly, the 18 elements of order 9 are:

(1,0)
(1,1)
(1,2)
(2,0)
(2,1)
(2,2)
(4,0)
(4,1)
(4,2)
(5,0)
(5,1)
(5,2)
(7,0)
(7,1)
(7,2)
(8,0)
(8,1)
(8,2)

for example:

(5,2) + (5,2) + (5,2) = (1,1) + (5,2) = (6,0), which is not the identity, so (5,2) is not of order 3. therefore (5,2) must be of order 9.

since i have explicitly listed 8 elements of order 3, and 18 elements of order 9, the only element missing is (0,0), the identity. clearly, looking at the number of elements of order 3 was faster.