Solve Group Theory Problem: (Z_4 x Z_4 x Z_8)/<(1,2,4)>

[ehrenfest]

[SOLVED] group theory problem

Homework Statement

Classify the factor group (Z_4 cross Z_4 cross Z_8)/<(1,2,4)> according to the fundamental theorem of finitely generated abelian groups.

Homework Equations

The Attempt at a Solution

<(1,2,4)> has order 4 so the factor group has order 32, so there are seven possibilities:

(Z_2)^5
Z_32
(Z_2)^3 cross Z_4
Z_16 cross Z_2
Z_8 cross Z_2 cross Z_2
Z_2 cross Z_4 cross Z_4
Z_8 cross Z_4

Anyone have any ideas about how to do this without doing a lot of tedious calculations?

 

[Hurkyl]

It seems evident to me: your relation says:

(1, 0, 0) + (0, 2, 0) + (0, 0, 4) = (0, 0, 0)

*shrug*


Maybe rewriting it as a quotient of Z³ would help?

 

[ehrenfest]

What is "it"?

 

[masnevets]

Let a,b,c be the generators of Z/4, Z/4, and Z/8, respectively. Then modding out by (1,2,4) is the same as saying that a=-2b-4c. So this means that a is completely determined by b and c, and hence doesn't matter anymore. Now the question is whether b and c can be whatever they want (between 0 and 3 and 0 and 7, respectively) and give different elements. I'll leave the rest to you.

 

[ehrenfest]

Let a,b,c be the generators of Z/4, Z/4, and Z/8, respectively. Then modding out by (1,2,4) is the same as saying that a=-2b-4c. So this means that a is completely determined by b and c, and hence doesn't matter anymore.

I am kind of confused about this. Does Z/4 mean the same thing as Z_4 i.e. the integers mod 4? If so, then a=b=c=1, and 1 is not equal to -6.

 

[ehrenfest]

Can someone please elaborate on what masnevets is saying?

 

[ehrenfest]

anyone? this problem is killing me!

 

[ehrenfest]

please?

 

[masnevets]

Yes, Z/4 is the integers modulo 4. Yes, 1 is not equal to -6, but you're confusing what I mean now. a, b, and c mean (1,0,0), (0,1,0), and (0,0,1), respectively.

 

[ehrenfest]

So, you're just saying that the class of (1,0,0) is in the same as the class of (0,-2,-4) ? I agree. Those two elements are clearly in the same coset. But how does that help you classify the quotient group according to the fundamental theorem of finitely generated abelian groups?!

 

[NateTG]

The natural homomorphism from the group to the qotient is going to be onto, so the image of a set of generators is a set of generators.

 

[ehrenfest]

The natural homomorphism from the group to the qotient is going to be onto, so the image of a set of generators is a set of generators.

I am not sure I have seen that proof...

But you're saying that the (1,0,0)+<(1,2,4)>,(0,1,0)+<(1,2,4)>,(0,0,1)+<(1,2,4)> will generate the quotient group?

Can someone just give me a concrete instruction so that I can make some progress classifying this group!

Maybe I need to use the Fundamental Homomorphism Theorem...to use this I need to find a group G' and a homomorphism phi such that ker(phi)=<(1,2,4)>. How would I figure out what G' is and what phi is though...

 

[ehrenfest]

anyone?

 

[ehrenfest]

anyone?