Thread: 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



(is this isomorphic to ?)


Now G cannot be , for G has no element of order 27.

I also think G cannot be , 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 , then

So I am left with . 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 , would have order 9, but I dont think there's 18 of those.

Thank you for your time.

It is easy to see that G is to for the order of an element of this group if the lcm(a,b), where has elements of order 9,3,1 (order of elements divides order of group, since cyclic these are the only 3 possible). So since is cyclic, if has generators, or elements of order 9. So basically 6 * 3, elements of 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.

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.