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

ianchenmu

Member
Feb 3, 2013
74
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? )
 

jakncoke

Active member
Jan 11, 2013
68
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 [tex](\mathbb{Z}/32\mathbb{Z}[/tex] 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) ?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
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 $C_2$.
 

Klaas van Aarsen

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Mar 5, 2012
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Note that there is more than one proper sub group with 4 elements.
 

ianchenmu

Member
Feb 3, 2013
74
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 [tex](\mathbb{Z}/32\mathbb{Z}[/tex] 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 $\mathbb{Z}/32\mathbb{Z}$
 

jakncoke

Active member
Jan 11, 2013
68

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
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 $\mathbb R^*$.
 

ianchenmu

Member
Feb 3, 2013
74
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?)
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,856
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.
 

ianchenmu

Member
Feb 3, 2013
74
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 $(\mathbb{Z}/32\mathbb{Z})^\times$?
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
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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.
 
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ianchenmu

Member
Feb 3, 2013
74
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?
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
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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?
 

ianchenmu

Member
Feb 3, 2013
74
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?
 

ianchenmu

Member
Feb 3, 2013
74
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 $(\mathbb{Z}/32\mathbb{Z})^\times$!
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
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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. :eek:)
 

ianchenmu

Member
Feb 3, 2013
74
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?
 

Klaas van Aarsen

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Mar 5, 2012
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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).
 
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jakncoke

Active member
Jan 11, 2013
68
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>
 

ianchenmu

Member
Feb 3, 2013
74
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)
 
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jakncoke

Active member
Jan 11, 2013
68
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 [tex]2^{4}[/tex] 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
 

ianchenmu

Member
Feb 3, 2013
74
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?
 

ianchenmu

Member
Feb 3, 2013
74
Since the order of every subgroup divides the order of the group [tex]2^{4}[/tex] 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 $<\bar{7},\bar{15}>$?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
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 $17^{-1}$.