Non-Isomorphic Abelian Groups of Order 54 and the Isomorphism of Factor Groups

[GatorPower]

Homework Statement

Determine how many non-isomorphic (and which) abelian groups there are of order 54.

Determine which of these groups the factor group Z6 x Z18 / <(3,0)> is isomorphic to.

Homework Equations

The Attempt at a Solution

Fundamental theorem for abelian groups gives:

54 = 2*3^3 and then the groups are

Z2 x Z3 x Z3 x Z3
Z2 x Z9 x Z3
Z2 x Z27


For the second part: <(3,0)> = { (0,0), (3,0) } so we get that the order of our group is (6*18)/2 = 54 as expected. But how do I decide which group it is isomorphic to? I have tried looking at cosets like (1,0) + <(3,0)> but don't seem to get anywhere...

 

[micromass]

You could ask yourself the following questions:

- Is my group cyclic?
- Do I have elements of order 9?

Answer these questions for the three groups you've found, and for the factor group in question. From the answer, you can discriminate between the groups...

 

[GatorPower]

The two first groups are not cyclic, only Z2 x Z27.

One does not have an elt of order 9 since 18/9 = 2 but <(3,0)> + (0,2) does not have order 9 seeing that (27,18) = (3,0)

Our group is not cyclic, and does not contain an elt of order 9 so hence it is isomorphic to the first group. Correct?

 

[micromass]

but <(3,0)> + (0,2) does not have order 9 seeing that (27,18) = (3,0)

I don't quite understand this part of your argument. What does (27,18)=(3,0) have to do with <(3,0)>+(0,2) and order 9?

 

[GatorPower]

I think I have thought about something wrong. There is in fact an element of order 9 which is (0,2) in the factor group. Then the factor group must be isomorphic to the SECOND one.

 

[micromass]

These elements do equal (0,0)+<(3,0)> when added 9 times! Note that

(27,18)+<(3,0)>=(0,18)+(27,0)+<(3,0)>=(0,18)+<(3,0)>=(0,0)+<(3,0)>...

 

[GatorPower]

So it is the second one? What kind of strategies does one use on this kind of problems? My book is pretty bad at explaining this..

 

[micromass]

Yes, it is the second one. For these kinds of problem, find some properties with discriminate between groups (like order, abelianness,...). Then check which of the properties the group in question has, and make your conclusions from that!

 

[GatorPower]

Another problem: Order is now 16 and the factor group is Z6 x Z8 / <(2,0)>.

(0,1) + <(2,0)> has order 8, so the only possible isomorphism is to Z8 x Z2 or Z16. But Z16 is cyclic, and the factor group is not so the isomorphism is to Z8 x Z2. Correct?

 

[micromass]

That seems to be correct!

 

[GatorPower]

Another question! Factor group Z6 x Z8 / <(2,0)>.

(0,1) + <(2,0)> has order 8, so we have two possible isomorphisms to Z8 x Z2 and Z16.
Seeing that the factor group don't have any elements with order 16 we decide it is isomorphic to Z8 x Z2. Correct?

 

[micromass]

Another question! Factor group Z6 x Z8 / <(2,0)>.

(0,1) + <(2,0)> has order 8, so we have two possible isomorphisms to Z8 x Z2 and Z16.
Seeing that the factor group don't have any elements with order 16 we decide it is isomorphic to Z8 x Z2. Correct?

Correct! Maybe you can make it easier on yourself and prove this result:

If N is a normal subgroup of G, then [tex]\frac{G\times H}{N\times \{0\}}\cong (G/N)\times H[/tex]. This would also solve all your exercises...

 

[GatorPower]

Correct! Maybe you can make it easier on yourself and prove this result:

If N is a normal subgroup of G, then [tex]\frac{G\times H}{N\times \{0\}}\cong (G/N)\times H[/tex]. This would also solve all your exercises...

Thank you =)