Identify isomorphism type for each proper subgroup of (Z/32Z)*

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Discussion Overview

The discussion revolves around identifying the isomorphism types of each proper subgroup of the group $(\mathbb{Z}/32\mathbb{Z})^{\times }$. Participants explore the definitions and implications of isomorphism types, the structure of the group, and the characteristics of its subgroups.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • Some participants seek clarification on what "isomorphism type" entails, questioning whether it involves listing isomorphic groups for each subgroup.
  • Others propose that the proper subgroups of $(\mathbb{Z}/32\mathbb{Z})^{\times }$ include those of orders 1, 2, 4, 8, and 16, suggesting they are isomorphic to cyclic groups $Z_2$, $Z_4$, $Z_8$, and $Z_{16}$.
  • There is a discussion about the elements of $(\mathbb{Z}/32\mathbb{Z})^{\times }$, with some participants noting that it consists of integers coprime to 32.
  • Some participants express the need to compute the subgroup generated by specific elements, such as $\langle \overline{3} \rangle$ and $\langle \overline{9} \rangle$, to identify subgroup relationships.
  • There are claims that $\langle \overline{3} \rangle$ generates the entire group, while others note that $\langle \overline{9} \rangle$ generates a proper subgroup.
  • Participants discuss the order of elements and their implications for subgroup structure, with some noting that groups of order 2 are cyclic and correspond to elements of order 2.
  • There is mention of groups of order 4 potentially being either $Z_2 \times Z_2$ or $Z_4$, with specific elements identified as generating these subgroups.
  • Further exploration includes the classification of groups of order 8 and their possible structures, emphasizing that they cannot be non-abelian.

Areas of Agreement / Disagreement

Participants express various viewpoints on subgroup structures and isomorphism types, with some agreement on the nature of certain subgroups, but no consensus on the complete classification or identification of all isomorphism types.

Contextual Notes

Participants note the complexity of identifying all subgroups and their isomorphism types, indicating that the task may involve considerable computation and exploration of subgroup relationships.

Who May Find This Useful

This discussion may be useful for those studying group theory, particularly in understanding the structure of multiplicative groups of integers modulo n and subgroup classification.

  • #31
ianchenmu said:
why $k$,$l$ only limit to 0,1,2,3, and 0,1...I think they can be any integer

The order of 7 is 4, so any higher $k$ than 3 will not yield a new element.
 
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  • #32
I like Serena said:
Well, did you try to repeatedly multiply each element with each element?

There are some patterns that you can use of course.
- Since your group is abelian, each element in your subgroup can be written as $7^k\cdot15^\ell$, where $k=0,1,2,3$ and $\ell=0,1$.
- Therefore the maximum order of the subgroup is 8.
- Since it contains at least $1, 7, 7^2, 7^3, 15$ the minimum order is 5.
- Since the order has to divide 16 (the order of the group), it has to be 8.
- A subgroup of order 8 must be isomorphic with either of $Z_8, \quad Z_4 \times Z_2, \quad Z_2 \times Z_2 \times Z_2$, which are the only possible groups of order 8.
- Since you have an element of order 4 and also a non-overlapping element of order 2, it has to be $Z_4 \times Z_2$, so you can predict what the multiplication table will look like.

But I've found and computed $<\bar{15},\bar{17},\bar{31}>$ only has four elements {$\bar{1},\bar{15},\bar{17},\bar{31}$}? why? it should be 8 elements since it's $\quad Z_2 \times Z_2 \times Z_2$, what's it's isomorphism type?
 
  • #33
ianchenmu said:
But I've found and computed $<\bar{15},\bar{17},\bar{31}>$ only has four elements {$\bar{1},\bar{15},\bar{17},\bar{31}$}? why? it should be 8 elements since it's $\quad Z_2 \times Z_2 \times Z_2$, what's it's isomorphism type?

Well, it isn't isomorphic with $\quad Z_2 \times Z_2 \times Z_2$.
It can't be since it only has 4 elements.
A group of 4 elements has to be isomorphic with either of $Z_4, \quad Z_2 \times Z_2, \quad V_4$.
Which group do you think it is isomorphic to?

See wiki for a list of small groups.
 
  • #34
I like Serena said:
Well, it isn't isomorphic with $\quad Z_2 \times Z_2 \times Z_2$.
It can't be since it only has 4 elements.
A group of 4 elements has to be isomorphic with either of $Z_4, \quad Z_2 \times Z_2, \quad V_4$.
Which group do you think it is isomorphic to?

See wiki for a list of small groups.

I am not sure what $<\bar{15},\bar{17},\bar{31}>$ is isomorphic to since it is generated by 3 elements ...and if it's isomorphic to $Z_4$ but $Z_4$ is generated by only one element. Can they be isomorphic?...I am not sure.
 
  • #35
ianchenmu said:
I am not sure what $<\bar{15},\bar{17},\bar{31}>$ is isomorphic to since it is generated by 3 elements ...and if it's isomorphic to $Z_4$ but $Z_4$ is generated by only one element. Can they be isomorphic?...I am not sure.

What is the order of each of the elements of $<\bar{15},\bar{17},\bar{31}>$?
And what is the order of each of the elements of $Z_4$?

Basically isomorphic means that all elements behave the same, have the same order, and have the same multiplication table.
 
  • #36
I like Serena said:
What is the order of each of the elements of $<\bar{15},\bar{17},\bar{31}>$?
And what is the order of each of the elements of $Z_4$?

Basically isomorphic means that all elements behave the same, have the same order, and have the same multiplication table.

So $<\bar{15},\bar{17},\bar{31}>$ is isomorphic to $Z_4\times Z_4$? since every element in $Z_4\times Z_4$ has order 2.

Can you tell me exactly what $<\bar{15},\bar{17},\bar{31}>$ is isomorphic to?
 
  • #37
ianchenmu said:
So $<\bar{15},\bar{17},\bar{31}>$ is isomorphic to $Z_4\times Z_4$? since every element in $Z_4\times Z_4$ has order 2.

It can't be, since $Z_4\times Z_4$ has order 16.

Can you tell me exactly what $<\bar{15},\bar{17},\bar{31}>$ is isomorphic to?

Okay, okay, it's isomorphic to $Z_2 \times Z_2$.
$\bar{1} \mapsto (0,0)$
$\bar{15} \mapsto (0,1)$
$\bar{17} \mapsto (1,0)$
$\bar{31} \mapsto (1,1)$

This matches since $\bar{15} \cdot \bar{15} = \bar 1$, while at the same time $(0,1) + (0,1) = (0,0) \mod 2$.
And we have $\bar{15} \cdot \bar{17} = \bar{31}$, while at the same time $(0,1) + (1,0) = (1,1) \mod 2$.
 
  • #38
Thank you
 
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