The discussion focuses on identifying the isomorphism types of each proper subgroup of the group $(\mathbb{Z}/32\mathbb{Z})^{\times}$. The proper subgroups identified include those of orders 1, 2, 4, 8, and 16, with specific isomorphism types such as $C_2$, $C_4$, $C_8$, and $C_{16}$. The subgroup generated by $\overline{3}$ is noted to generate the entire group, while other elements like $\overline{9}$ generate proper subgroups. The discussion also clarifies that the notation $(\mathbb{Z}/32\mathbb{Z})^{\times}$ represents the multiplicative group of integers modulo 32, excluding non-invertible elements.
PREREQUISITESMathematicians, particularly those specializing in abstract algebra, students studying group theory, and anyone interested in the properties of finite groups and their substructures.
ianchenmu said:why $k$,$l$ only limit to 0,1,2,3, and 0,1...I think they can be any integer
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.
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?
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.
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.
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.
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.
Can you tell me exactly what $<\bar{15},\bar{17},\bar{31}>$ is isomorphic to?