List the subgroups of $mathbb{Z}_{26}$ [closed]

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I think the answer is $mathbb{Z}_n$ where $n$ is all the factors of $26$. Is this right?

abstract-algebra group-theory

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edited Nov 22 '18 at 9:50

user26857

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asked Oct 3 '16 at 11:54

PolkaDotPolkaDot

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closed as off-topic by José Carlos Santos, Gibbs, amWhy, ancientmathematician, Rebellos Nov 22 '18 at 16:01

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0

I think the answer is $mathbb{Z}_n$ where $n$ is all the factors of $26$. Is this right?

abstract-algebra group-theory

share|cite|improve this question

edited Nov 22 '18 at 9:50

user26857

39.3k124083

asked Oct 3 '16 at 11:54

PolkaDotPolkaDot

245

closed as off-topic by José Carlos Santos, Gibbs, amWhy, ancientmathematician, Rebellos Nov 22 '18 at 16:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Gibbs, amWhy, ancientmathematician, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

0

0

0

I think the answer is $mathbb{Z}_n$ where $n$ is all the factors of $26$. Is this right?

abstract-algebra group-theory

share|cite|improve this question

edited Nov 22 '18 at 9:50

user26857

39.3k124083

asked Oct 3 '16 at 11:54

PolkaDotPolkaDot

245

I think the answer is $mathbb{Z}_n$ where $n$ is all the factors of $26$. Is this right?

abstract-algebra group-theory

abstract-algebra group-theory

share|cite|improve this question

edited Nov 22 '18 at 9:50

user26857

39.3k124083

asked Oct 3 '16 at 11:54

PolkaDotPolkaDot

245

share|cite|improve this question

edited Nov 22 '18 at 9:50

user26857

39.3k124083

asked Oct 3 '16 at 11:54

PolkaDotPolkaDot

245

share|cite|improve this question

share|cite|improve this question

edited Nov 22 '18 at 9:50

user26857

39.3k124083

edited Nov 22 '18 at 9:50

user26857

39.3k124083

edited Nov 22 '18 at 9:50

user26857

39.3k124083

39.3k124083

asked Oct 3 '16 at 11:54

PolkaDotPolkaDot

245

asked Oct 3 '16 at 11:54

PolkaDotPolkaDot

245

asked Oct 3 '16 at 11:54

PolkaDotPolkaDot

245

245

closed as off-topic by José Carlos Santos, Gibbs, amWhy, ancientmathematician, Rebellos Nov 22 '18 at 16:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Gibbs, amWhy, ancientmathematician, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, Gibbs, amWhy, ancientmathematician, Rebellos Nov 22 '18 at 16:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Gibbs, amWhy, ancientmathematician, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.

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2 Answers
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1

$26 = 2cdot 13$. Both 2 and 13 are the prime factors of $26$, and cannot be reduced further.

So we know that So $$mathbb Z_{26} equiv mathbb Z_2 times mathbb Z_{13}$$

We have Lagrange Theorem to thank for proving that:

"...for any finite group $G,$ the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G.$ The converse of the theorem applies only to cyclic groups.

More to the point: You need to prove that these are, and are the only, subgroups of $mathbb Z_{26}$:

$mathbb Z_{26},; {0},; {0,13},$ and ${0, 2,4,6,8,10, 12, 14, 16, 18, 20, 22, 24}.$

Notice the orders of of the groups listed: in order, we have $26, 1, 2, 13 $, which all divide $26$ (with no remainder).

And, indeed, we have that a cyclic group of order $n$ is isomorphic to the group $mathbb Z_n$

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edited Oct 3 '16 at 12:40

answered Oct 3 '16 at 12:09

amWhyamWhy

192k28225439

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1

Not quite: Every subgroup of $mathbb{Z}_{26}$ is isomorphic to $mathbb{Z}_n$ for $n|26$, but such a $mathbb{Z}_n$ is not contained in $mathbb{Z}_{26}$ in a strict sense. The list of subgroups is the following:

• trivial subgroups: $mathbb{Z}_{26}$, ${0}$


• non-trivial subgroups: ${0,2,4,6,8,10,12,14,16,18,20,22,24}$, ${0,13}$.

Can you prove by yourself that these are indeed all subgroups? Do you see why the first non-trivial subgroup is isomorphic to $mathbb{Z}_{13}$ and the second one is isomorphic to $mathbb{Z}_2$?

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answered Oct 3 '16 at 12:07

JanikJanik

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$26 = 2cdot 13$. Both 2 and 13 are the prime factors of $26$, and cannot be reduced further.

So we know that So $$mathbb Z_{26} equiv mathbb Z_2 times mathbb Z_{13}$$

We have Lagrange Theorem to thank for proving that:

"...for any finite group $G,$ the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G.$ The converse of the theorem applies only to cyclic groups.

More to the point: You need to prove that these are, and are the only, subgroups of $mathbb Z_{26}$:

$mathbb Z_{26},; {0},; {0,13},$ and ${0, 2,4,6,8,10, 12, 14, 16, 18, 20, 22, 24}.$

Notice the orders of of the groups listed: in order, we have $26, 1, 2, 13 $, which all divide $26$ (with no remainder).

And, indeed, we have that a cyclic group of order $n$ is isomorphic to the group $mathbb Z_n$

share|cite|improve this answer

edited Oct 3 '16 at 12:40

answered Oct 3 '16 at 12:09

amWhyamWhy

192k28225439

add a comment | 

1

$26 = 2cdot 13$. Both 2 and 13 are the prime factors of $26$, and cannot be reduced further.

So we know that So $$mathbb Z_{26} equiv mathbb Z_2 times mathbb Z_{13}$$

We have Lagrange Theorem to thank for proving that:

"...for any finite group $G,$ the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G.$ The converse of the theorem applies only to cyclic groups.

More to the point: You need to prove that these are, and are the only, subgroups of $mathbb Z_{26}$:

$mathbb Z_{26},; {0},; {0,13},$ and ${0, 2,4,6,8,10, 12, 14, 16, 18, 20, 22, 24}.$

Notice the orders of of the groups listed: in order, we have $26, 1, 2, 13 $, which all divide $26$ (with no remainder).

And, indeed, we have that a cyclic group of order $n$ is isomorphic to the group $mathbb Z_n$

share|cite|improve this answer

edited Oct 3 '16 at 12:40

answered Oct 3 '16 at 12:09

amWhyamWhy

192k28225439

add a comment | 

1

1

1

$26 = 2cdot 13$. Both 2 and 13 are the prime factors of $26$, and cannot be reduced further.

So we know that So $$mathbb Z_{26} equiv mathbb Z_2 times mathbb Z_{13}$$

We have Lagrange Theorem to thank for proving that:

"...for any finite group $G,$ the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G.$ The converse of the theorem applies only to cyclic groups.

More to the point: You need to prove that these are, and are the only, subgroups of $mathbb Z_{26}$:

$mathbb Z_{26},; {0},; {0,13},$ and ${0, 2,4,6,8,10, 12, 14, 16, 18, 20, 22, 24}.$

Notice the orders of of the groups listed: in order, we have $26, 1, 2, 13 $, which all divide $26$ (with no remainder).

And, indeed, we have that a cyclic group of order $n$ is isomorphic to the group $mathbb Z_n$

share|cite|improve this answer

edited Oct 3 '16 at 12:40

answered Oct 3 '16 at 12:09

amWhyamWhy

192k28225439

$26 = 2cdot 13$. Both 2 and 13 are the prime factors of $26$, and cannot be reduced further.

So we know that So $$mathbb Z_{26} equiv mathbb Z_2 times mathbb Z_{13}$$

We have Lagrange Theorem to thank for proving that:

"...for any finite group $G,$ the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G.$ The converse of the theorem applies only to cyclic groups.

More to the point: You need to prove that these are, and are the only, subgroups of $mathbb Z_{26}$:

$mathbb Z_{26},; {0},; {0,13},$ and ${0, 2,4,6,8,10, 12, 14, 16, 18, 20, 22, 24}.$

Notice the orders of of the groups listed: in order, we have $26, 1, 2, 13 $, which all divide $26$ (with no remainder).

And, indeed, we have that a cyclic group of order $n$ is isomorphic to the group $mathbb Z_n$

share|cite|improve this answer

edited Oct 3 '16 at 12:40

answered Oct 3 '16 at 12:09

amWhyamWhy

192k28225439

share|cite|improve this answer

share|cite|improve this answer

edited Oct 3 '16 at 12:40

edited Oct 3 '16 at 12:40

edited Oct 3 '16 at 12:40

answered Oct 3 '16 at 12:09

amWhyamWhy

192k28225439

answered Oct 3 '16 at 12:09

amWhyamWhy

192k28225439

answered Oct 3 '16 at 12:09

amWhyamWhy

192k28225439

192k28225439

add a comment | 

add a comment | 

1

Not quite: Every subgroup of $mathbb{Z}_{26}$ is isomorphic to $mathbb{Z}_n$ for $n|26$, but such a $mathbb{Z}_n$ is not contained in $mathbb{Z}_{26}$ in a strict sense. The list of subgroups is the following:

• trivial subgroups: $mathbb{Z}_{26}$, ${0}$


• non-trivial subgroups: ${0,2,4,6,8,10,12,14,16,18,20,22,24}$, ${0,13}$.

Can you prove by yourself that these are indeed all subgroups? Do you see why the first non-trivial subgroup is isomorphic to $mathbb{Z}_{13}$ and the second one is isomorphic to $mathbb{Z}_2$?

share|cite|improve this answer

answered Oct 3 '16 at 12:07

JanikJanik

1,4602418

add a comment | 

1

Not quite: Every subgroup of $mathbb{Z}_{26}$ is isomorphic to $mathbb{Z}_n$ for $n|26$, but such a $mathbb{Z}_n$ is not contained in $mathbb{Z}_{26}$ in a strict sense. The list of subgroups is the following:

• trivial subgroups: $mathbb{Z}_{26}$, ${0}$


• non-trivial subgroups: ${0,2,4,6,8,10,12,14,16,18,20,22,24}$, ${0,13}$.

Can you prove by yourself that these are indeed all subgroups? Do you see why the first non-trivial subgroup is isomorphic to $mathbb{Z}_{13}$ and the second one is isomorphic to $mathbb{Z}_2$?

share|cite|improve this answer

answered Oct 3 '16 at 12:07

JanikJanik

1,4602418

add a comment | 

1

1

1

Not quite: Every subgroup of $mathbb{Z}_{26}$ is isomorphic to $mathbb{Z}_n$ for $n|26$, but such a $mathbb{Z}_n$ is not contained in $mathbb{Z}_{26}$ in a strict sense. The list of subgroups is the following:

• trivial subgroups: $mathbb{Z}_{26}$, ${0}$


• non-trivial subgroups: ${0,2,4,6,8,10,12,14,16,18,20,22,24}$, ${0,13}$.

Can you prove by yourself that these are indeed all subgroups? Do you see why the first non-trivial subgroup is isomorphic to $mathbb{Z}_{13}$ and the second one is isomorphic to $mathbb{Z}_2$?

share|cite|improve this answer

answered Oct 3 '16 at 12:07

JanikJanik

1,4602418

Not quite: Every subgroup of $mathbb{Z}_{26}$ is isomorphic to $mathbb{Z}_n$ for $n|26$, but such a $mathbb{Z}_n$ is not contained in $mathbb{Z}_{26}$ in a strict sense. The list of subgroups is the following:

• trivial subgroups: $mathbb{Z}_{26}$, ${0}$


• non-trivial subgroups: ${0,2,4,6,8,10,12,14,16,18,20,22,24}$, ${0,13}$.

Can you prove by yourself that these are indeed all subgroups? Do you see why the first non-trivial subgroup is isomorphic to $mathbb{Z}_{13}$ and the second one is isomorphic to $mathbb{Z}_2$?

share|cite|improve this answer

answered Oct 3 '16 at 12:07

JanikJanik

1,4602418

share|cite|improve this answer

share|cite|improve this answer

answered Oct 3 '16 at 12:07

JanikJanik

1,4602418

answered Oct 3 '16 at 12:07

JanikJanik

1,4602418

answered Oct 3 '16 at 12:07

JanikJanik

1,4602418

1,4602418

add a comment | 

add a comment | 

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