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

[I like Serena]

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.

 

[i_a_n]

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?

 

[I like Serena]

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.

 

[i_a_n]

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.

 

[I like Serena]

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.

 

[i_a_n]

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?

 

[I like Serena]

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 $(0,1) + (1,0) = (1,1) \mod 2$.

 

[i_a_n]

Thank you