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

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SUMMARY

The factor group (Z_4 x Z_4 x Z_8)/<(1,2,4)> has been classified as having an order of 32, leading to seven potential structures according to the fundamental theorem of finitely generated abelian groups. The relevant structures include (Z_2)^5, Z_32, (Z_2)^3 x Z_4, Z_16 x Z_2, Z_8 x Z_2 x Z_2, Z_2 x Z_4 x Z_4, and Z_8 x Z_4. The discussion emphasizes the importance of understanding the relationship between the generators of the groups and the implications of the quotient operation. Participants suggest using the Fundamental Homomorphism Theorem to further analyze the group structure.

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  • Familiarity with the fundamental theorem of finitely generated abelian groups.
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  • Study the Fundamental Homomorphism Theorem in detail to apply it to group classifications.
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[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?
 
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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?
 
What is "it"?
 
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.
 
masnevets said:
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.
 
Can someone please elaborate on what masnevets is saying?
 
anyone? this problem is killing me!
 
please?
 
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.
 
  • #10
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?!
 
  • #11
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.
 
  • #12
NateTG said:
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...
 
Last edited:
  • #13
anyone?
 
  • #14
anyone?
 

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