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

[i_a_n]

The question is to identify isomorphism type for each proper subgroup of $(\mathbb{Z}/32\mathbb{Z})^{\times }$.
(what's the "isomorphism type" means? Does the question mean we need to list all the ismorphism between of each subgroup and the respectively another group that is isomorphic to the subgroup? If so, what are they then? )

 

[jakncoke1]

All i found regarding type was

if G is a finite group direct products of cyclic groups of order ${p_1}^{r_1},...,{p_k}^{r_k}$ with $p_i \leq p_k$ for $i<k$. (prime power cyclic group direct product)

The type was defined to be the k tuple $(p_1)^{r_1}, ...,(p_k)^{r_k})$.

I guess for $$(\mathbb{Z}/32\mathbb{Z}$$ the proper subgroups are of order 1,2,4,8,16. Since cyclic. Subgroups are isomorphic to $Z_{2},Z_{4},Z_{8},Z_{16}$ so i gess they are of type, (2), (4), (8), (16) ?

 

[I like Serena]

The question is to identify isomorphism type for each proper subgroup of $(\mathbb{Z}/32\mathbb{Z})^{\times }$.

(what's the "isomorphism type" means? Does the question mean we need to list all the ismorphism between of each subgroup and the respectively another group that is isomorphic to the subgroup? If so, what are they then? )

I believe that for each proper subgroup you need to identify a group that it is isomorphic to.
For instance the sub group with 2 elements {1,17} is isomorphic to $C_2$.

 

[I like Serena]

Note that there is more than one proper sub group with 4 elements.

 

[i_a_n]

All i found regarding type was

if G is a finite group direct products of cyclic groups of order ${p_1}^{r_1},...,{p_k}^{r_k}$ with $p_i \leq p_k$ for $i<k$. (prime power cyclic group direct product)

The type was defined to be the k tuple $(p_1)^{r_1}, ...,(p_k)^{r_k})$.

I guess for $$(\mathbb{Z}/32\mathbb{Z}$$ the proper subgroups are of order 1,2,4,8,16. Since cyclic. Subgroups are isomorphic to $Z_{2},Z_{4},Z_{8},Z_{16}$ so i gess they are of type, (2), (4), (8), (16) ?

It's $(\mathbb{Z}/32\mathbb{Z})^{\times }$, not $\mathbb{Z}/32\mathbb{Z}$

 

[jakncoke1]

It's $(\mathbb{Z}/32\mathbb{Z})^{\times }$, not $\mathbb{Z}/32\mathbb{Z}$

can you tell me what the cross represents?

 

[I like Serena]

can you tell me what the cross represents?

The cross represents "times".
It's the set of whole numbers mod 32 with $\times$ as operation.
In particular every element that does not have an inverse is removed from the set.
In other words $(\mathbb Z/32\mathbb Z)^\times = (\{1, 3, 5, ..., 31\}, \times)$.
It is also denoted as $(\mathbb Z/32\mathbb Z)^*$.
You may be more familiar with for instance $\mathbb R^*$.

 

[i_a_n]

can you tell me what the cross represents?

The question is to draw the complete lattice of subgroups of $(\mathbb{Z}/32\mathbb{Z})^{\times }$ and for each proper subgroup, identify the isomorphism type.(Accoding to definition,
$(\mathbb{Z}/n\mathbb{Z})^{\times }=\left \{ {\bar{a}\in \mathbb{Z}/n\mathbb{Z}|(a,n)=1}\right \}$
so $(\mathbb{Z}/32\mathbb{Z})^{\times }=\left \{ {\overline{1},\overline{3},\overline{5},\overline{7},\overline{9},...,\overline{31}}\right \}$, but what then? Can these elements form any subgroup? how to draw the lattice and for each proper subgroup, identify the isomorphism type?)

 

[I like Serena]

so $(\mathbb{Z}/32\mathbb{Z})^{\times }=\left \{ {\overline{1},\overline{3},\overline{5},\overline{7},\overline{9},...,\overline{31}}\right \}$, but what then? Can these elements form any subgroup? how to draw the lattice and for each proper subgroup, identify the isomorphism type?)

Which sub group is generated by $\langle \overline{3} \rangle$?
And what is $\langle \overline{9} \rangle$?
And...?

To identify each subgroup, start with 1 element and see what it generates.
If there is still space, try to add a 2nd element and see what it generates.

 

[i_a_n]

Which sub group is generated by $\langle \overline{3} \rangle$?
And what is $\langle \overline{9} \rangle$?
And...?

To identify each subgroup, start with 1 element and see what it generates.
If there is still space, try to add a 2nd element and see what it generates.

But $\langle \overline{3} \rangle$={$\overline{3}$,$ \overline{9} $,$ \overline{27} $,$ \overline{17} $...}and $\langle \overline{3} \rangle$ includes every element in $(\mathbb{Z}/32\mathbb{Z})^\times$.

How can we draw a lattice of subgroups of $(\mathbb{Z}/32\mathbb{Z})^\times$?

 

[I like Serena]

But $\langle \overline{3} \rangle$={$\overline{3}$,$ \overline{9} $,$ \overline{27} $,$ \overline{17} $...}and $\langle \overline{3} \rangle$ includes every element in $(\mathbb{Z}/32\mathbb{Z})^\times$.

How can we draw a lattice of subgroups of $(\mathbb{Z}/32\mathbb{Z})^\times$?

Yes. $\langle \overline{3} \rangle$ generates the entire group.
But then $\overline{3}\cdot \overline{3} = \overline{9}$ won't...

Edit: sorry, that's not true. See below.

 

[i_a_n]

But $\langle \overline{3} \rangle$={$\overline{3}$,$ \overline{9} $,$ \overline{27} $,$ \overline{17} $...}and $\langle \overline{3} \rangle$ includes every element in $(\mathbb{Z}/32\mathbb{Z})^\times$.

How can we draw a lattice of subgroups of $(\mathbb{Z}/32\mathbb{Z})^\times$?

Oh I got what you mean, $\langle \overline{9} \rangle$ has less element. and so on... so there exists subgroup relationships. Is that right?

 

[I like Serena]

Oh I got what you mean, $\langle \overline{9} \rangle$ has less element. and so on... so there exists subgroup relationships. Is that right?

Correct!
So $\overline{9}$ is the first element that generates a proper subgroup.
How big is it?
To which group is it isomorphic?

 

[i_a_n]

Correct!
So $\overline{9}$ is the first element that generates a proper subgroup.
How big is it?
To which group is it isomorphic?

So do I need to compute $<\overline{a}>$ of each element $\overline{a}$ in $(\mathbb{Z}/32\mathbb{Z})^\times$ in order to get the lattice?...it's so much work. And I can't find the isomorphism type. And how to find this?

 

[i_a_n]

Yes. $\langle \overline{3} \rangle$ generates the entire group.
But then $\overline{3}\cdot \overline{3} = \overline{9}$ won't...

Wait! I computed that $\overline{3}$ has order 8 in $(\mathbb{Z}/32\mathbb{Z})^\times$!

 

[I like Serena]

So do I need to compute $<\overline{a}>$ of each element $\overline{a}$ in $(\mathbb{Z}/32\mathbb{Z})^\times$ in order to get the lattice?...it's so much work. And I can't find the isomorphism type. And how to find this?

Did you know that a cyclic group is a group that can be generated by 1 element?
Btw, a lot of the elements will be equivalent to another one.
For instance $9^{-1}$ will generate the same group as $9$.

Wait! I computed that $\overline{3}$ has order 8 in $(\mathbb{Z}/32\mathbb{Z})^\times$!

Right!
(I overlooked that myself. )

 

[i_a_n]

Did you know that a cyclic group is a group that can be generated by 1 element?
Btw, a lot of the elements will be equivalent to another one.
For instance $9^{-1}$ will generate the same group as $9$.

So after computing, I found $<\bar{3}> =<\bar{11}>=<\bar{19}>=<\bar{27}>$. So can I say this is the isomorphism type? What they are isomorphic to?

 

[I like Serena]

So after computing, I found $<\bar{3}> =<\bar{11}>=<\bar{19}>=<\bar{27}>$. So can I say this is the isomorphism type?

Good!
All of these groups have 8 elements.
Moreover, they are generated by 1 element.
This means that their isomorphism type is $C_8$ (or $Z_8$ or $\mathbb Z/8\mathbb Z$ depending on which notation you prefer).

 

[jakncoke1]

Ok.

Groups of order 2 are cyclic so the only subgroups of order 2 correspond to elements of order 2.

<15>,<17>,<31>

Groups of order 4 can be either $Z_2 \times Z_2$ or $Z_4$
$Z_4$ corresponds to cyclic elements of order 4.
so <7>,<9>,<23>,<25>
Since <7> = <23>
<9> = <25>
<7>,<9> are our distinct subgroups isomoprhic to $Z_4$.

$Z_2 \times Z_2$ corresponds to the direct product of subgroups generated by elements of order 2.

So <15>$\times$<17>
<15>$\times$<31>
<31>$\times$<17>

Now Groups of order 8 can be either $Z_8$, $Z_4 \times Z_2$, or $Z_2\times Z_2 \times Z_2$. It cannot be quaternions or $D_8$ because that group is not abelian while ours clearly is.

so $Z_8$ = <3>,<11>,<19>,<27>
$Z_4 \times Z_2$ is the <7> $\times$ <15>
<9> $\times$ <15>
<7> $\times$ <17>
<9> $\times$ <17>
<7> $\times$ <31>
<9> $\times$ <31>
so lastly $Z_2 \times Z_2 \times Z_2$
basically <15> $\times$ <17> $\times$ <31>

 

[i_a_n]

I am considering will any two , or three (or four possibly...)elements from $\bar{1},\bar{3},\bar{5},\bar{7}...\bar{31}$ will generate a subgroup of order m, m<32. Is this possible?And what's the order of the formed groups can be? As I searched, {$\bar{3},\bar{31}$} generates $(\mathbb{Z}/32\mathbb{Z})^\times $(see this page: Multiplicative group of integers modulo n - Wikipedia, the free encyclopedia)

 

[jakncoke1]

I am considering will any two , or three (or four possibly...)elements from $\bar{1},\bar{3},\bar{5},\bar{7}...\bar{31}$ will generate a subgroup of order m, m<32. Is this possible?And what's the order of the formed groups can be? As I searched, {$\bar{3},\bar{31}$} generates $(\mathbb{Z}/32\mathbb{Z})^\times $(see this page: Multiplicative group of integers modulo n - Wikipedia, the free encyclopedia)

Since the order of every subgroup divides the order of the group $$2^{4}$$ the only possible orders are 1,2,4,8,16 for subgroup sizes. I'm not sure if you are asking about generating sets of groups.

Generating set of a group - Wikipedia, the free encyclopedia

 

[i_a_n]

Ok.

Groups of order 2 are cyclic so the only subgroups of order 2 correspond to elements of order 2.

<15>,<17>,<31>

Groups of order 4 can be either $Z_2 \times Z_2$ or $Z_4$
$Z_4$ corresponds to cyclic elements of order 4.
so <7>,<9>,<23>,<25>
Since <7> = <23>
<9> = <25>
<7>,<9> are our distinct subgroups isomoprhic to $Z_4$.

$Z_2 \times Z_2$ corresponds to the direct product of subgroups generated by elements of order 2.

So <15>$\times$<17>
<15>$\times$<31>
<31>$\times$<17>

Now Groups of order 8 can be either $Z_8$, $Z_4 \times Z_2$, or $Z_2\times Z_2 \times Z_2$. It cannot be quaternions or $D_8$ because that group is not abelian while ours clearly is.

so $Z_8$ = <3>,<11>,<19>,<27>
$Z_4 \times Z_2$ is the <7> $\times$ <15>
<9> $\times$ <15>
<7> $\times$ <17>
<9> $\times$ <17>
<7> $\times$ <31>
<9> $\times$ <31>
so lastly $Z_2 \times Z_2 \times Z_2$
basically <15> $\times$ <17> $\times$ <31>

What $\times$ here means? Is that the subgroup is generated by the two element before and after $\times$? And what about subgroup of order 16?

 

[i_a_n]

Since the order of every subgroup divides the order of the group $$2^{4}$$ the only possible orders are 1,2,4,8,16 for subgroup sizes. I'm not sure if you are asking about generating sets of groups.

Generating set of a group - Wikipedia, the free encyclopedia

I mean, how could {$\bar{3},\bar{31}$} generate $\mathbb{Z}/32\mathbb{Z}^\times$? And for example, what's $<\bar{15},\bar{17}>$ and what's $<\bar{7},\bar{15}>$?

 

[I like Serena]

What $\times$ here means? Is that the subgroup is generated by the two element before and after $\times$? And what about subgroup of order 16?

The symbol $\times$ between sets is the so called cartesian product.
It forms a new set.
Its elements are ordered pairs.
For instance $Z_3 \times Z_2$ is the set {(0,0), (0,1), (1,0), (1,1), (2,0), (2,1)}.
Adding 2 elements means adding each component of an element separately.

The order of $(\mathbb Z / 32 \mathbb Z)^\times$ is 16.
The subgroup of order 16 is not a proper subgroup which your problem statement required.

 

[I like Serena]

I mean, how could {$\bar{3},\bar{31}$} generate $\mathbb{Z}/32\mathbb{Z}^\times$? And for example, what's $<\bar{15},\bar{17}>$ and what's $<\bar{7},\bar{15}>$?

I do not understand your question.

The group $<\bar{15},\bar{17}>$ is the group generated by 15 and 17.
It contains $15$ and $17$ and it also contains each element when multiplied (repeatedly) by either $15$, $17$, $15^{-1}$, or $17^{-1}$.

 

[i_a_n]

I do not understand your question.

The group $<\bar{15},\bar{17}>$ is the group generated by 15 and 17.
It contains $15$ and $17$ and it also contains each element when multiplied (repeatedly) by either $15$, $17$, $15^{-1}$, or $17^{-1}$.

I mean, since we know, $<\bar{15}>$={$<\bar{1}>,<\bar{15}>$},$<\bar{17}>$={$<\bar{1}>,<\bar{17}>$}, then $<\bar{15},\bar{17}>$={$<\bar{1}>,<\bar{15}>,<\bar{17}>$}? But how could it is of order 3?
And the same, since $<\bar{7}>$={$<\bar{1}>,<\bar{7}>,<\bar{17}>,<\bar{23}>$}, then $<\bar{7},\bar{15}>$={$<\bar{1}>,<\bar{7}>,<\bar{15}>,<\bar{17}>,<\bar{23}>$}...like it's of order 5...

So these conclusions should be not right.

My question is how to determine what elements in the subgroup generated by two (or three, four...) elements in $(\mathbb Z / 32 \mathbb Z)^\times$?

 

[I like Serena]

I mean, since we know, $<\bar{15}>$={$<\bar{1}>,<\bar{15}>$},$<\bar{17}>$={$<\bar{1}>,<\bar{17}>$}, then $<\bar{15},\bar{17}>$={$<\bar{1}>,<\bar{15}>,<\bar{17}>$}? But how could it is of order 3?
And the same, since $<\bar{7}>$={$<\bar{1}>,<\bar{7}>,<\bar{17}>,<\bar{23}>$}, then $<\bar{7},\bar{15}>$={$<\bar{1}>,<\bar{7}>,<\bar{15}>,<\bar{17}>,<\bar{23}>$}...like it's of order 5...

So these conclusions should be not right.

My question is how to determine what elements in the subgroup generated by two (or three, four...) elements in $(\mathbb Z / 32 \mathbb Z)^\times$?

What you write is not quite correct.

$<\bar{15}> = \{\bar{1},\bar{15}\}$

$<\bar{15},\bar{17}> = \{\bar{1},\bar{15},\bar{17},\bar{31} \}$ since $\bar{15} \cdot \bar{17} = \bar{31}$.
If we try to multiply each element with a second element we won't find new elements, so this is the complete subgroup.
It is isomorphic with $Z_2 \times Z_2 = (\{(0,0), (0,1), (1,0), (1,1)\}, +)$.

 

[i_a_n]

What you write is not quite correct.

$<\bar{15}> = \{\bar{1},\bar{15}\}$

$<\bar{15},\bar{17}> = \{\bar{1},\bar{15},\bar{17},\bar{31} \}$ since $\bar{15} \cdot \bar{17} = \bar{31}$.
If we try to multiply each element with a second element we won't find new elements, so this is the complete subgroup.
It is isomorphic with $Z_2 \times Z_2 = (\{(0,0), (0,1), (1,0), (1,1)\}, +)$.

But how to determine elements in $<\bar{7},\bar{15}>$? there should be 8 elements, right?

 

[I like Serena]

But how to determine elements in $<\bar{7},\bar{15}>$? there should be 8 elements, right?

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.

 

[i_a_n]

- 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 $\ell=0,1$.

why $k$,$l$ only limit to 0,1,2,3, and 0,1...I think they can be any integer

 

[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.