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- #1

(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? )

- Thread starter ianchenmu
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- Thread starter
- #1

(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? )

- Jan 11, 2013

- 68

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) ?

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- #3

- Mar 5, 2012

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I believe that for each proper subgroup you need to identify a group that it is isomorphic to.

(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? )

For instance the sub group with 2 elements {1,17} is isomorphic to $C_2$.

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- #4

- Mar 5, 2012

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

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- #5

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

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) ?

- Jan 11, 2013

- 68

can you tell me what the cross represents?It's $(\mathbb{Z}/32\mathbb{Z})^{\times }$, not $\mathbb{Z}/32\mathbb{Z}$

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- #7

- Mar 5, 2012

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The cross represents "times".can you tell me what the cross represents?

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 $\mathbb R^*$.

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- #8

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?)

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- #9

- Mar 5, 2012

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Which sub group is generated by $\langle \overline{3} \rangle$?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?)

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.

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- #10

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

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

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- #11

- Mar 5, 2012

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Yes. $\langle \overline{3} \rangle$ generates the entire group.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 $(\mathbb{Z}/32\mathbb{Z})^\times$?

But then $\overline{3}\cdot \overline{3} = \overline{9}$ won't...

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

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- #12

Oh I got what you mean, $\langle \overline{9} \rangle$ has less element. and so on... so there exists subgroup relationships. Is that right?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 $(\mathbb{Z}/32\mathbb{Z})^\times$?

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- #13

- Mar 5, 2012

- 8,856

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

So $\overline{9}$ is the first element that generates a proper subgroup.

How big is it?

To which group is it isomorphic?

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- #14

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?Correct!

So $\overline{9}$ is the first element that generates a proper subgroup.

How big is it?

To which group is it isomorphic?

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- #15

Wait! I computed that $\overline{3}$ has order 8 in $(\mathbb{Z}/32\mathbb{Z})^\times$!Yes. $\langle \overline{3} \rangle$ generates the entire group.

But then $\overline{3}\cdot \overline{3} = \overline{9}$ won't...

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- #16

- Mar 5, 2012

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Did you know that a cyclic group is a group that can be generated by 1 element?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?

Btw, a lot of the elements will be equivalent to another one.

For instance $9^{-1}$ will generate the same group as $9$.

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

(I overlooked that myself. )

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- #17

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?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 $9$.

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- #18

- Mar 5, 2012

- 8,856

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

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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- Jan 11, 2013

- 68

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>

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- #20

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)

(see this page: Multiplicative group of integers modulo n - Wikipedia, the free encyclopedia)

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- Jan 11, 2013

- 68

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.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)

Generating set of a group - Wikipedia, the free encyclopedia

- Thread starter
- #22

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?

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>

- Thread starter
- #23

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 $<\bar{7},\bar{15}>$?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

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- #24

- Mar 5, 2012

- 8,856

The symbol $\times$ between sets is the so calledWhat $\times$ here means? Is that the subgroup is generated by the two element before and after $\times$? And what about subgroup of order 16?

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

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- #25

- Mar 5, 2012

- 8,856

I do not understand your question.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 $<\bar{7},\bar{15}>$?

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 $17^{-1}$.