Which of the following is isomorphic to the group of units of $mathbb{Z}

Which of the following is isomorphic to the group of units of $mathbb{Z}_{35}$? [closed]

-1

$C_2 times C_2 times C_6$

$C_2 times C_{12}$

$C_{24}$

The group of units of $mathbb{Z}_{35}$ is $mathbb{Z}_{35}^#=
left{ 1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34 right}$.

asked Nov 29 '18 at 17:55

vladr10

closed as off-topic by Derek Holt, Jyrki Lahtonen, DRF, amWhy, ancientmathematician Dec 5 '18 at 17:48

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, DRF, amWhy, ancientmathematician

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

1

$begingroup$
what have you tried ? Why makes the three possibilities you gave non-isomorphic ?
$endgroup$
– user120527
Nov 29 '18 at 18:03

add a comment |

-1

$C_2 times C_2 times C_6$

$C_2 times C_{12}$

$C_{24}$

The group of units of $mathbb{Z}_{35}$ is $mathbb{Z}_{35}^#=
left{ 1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34 right}$.

asked Nov 29 '18 at 17:55

vladr10

closed as off-topic by Derek Holt, Jyrki Lahtonen, DRF, amWhy, ancientmathematician Dec 5 '18 at 17:48

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, DRF, amWhy, ancientmathematician

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

1

$begingroup$
what have you tried ? Why makes the three possibilities you gave non-isomorphic ?
$endgroup$
– user120527
Nov 29 '18 at 18:03

add a comment |

-1

$C_2 times C_2 times C_6$

$C_2 times C_{12}$

$C_{24}$

The group of units of $mathbb{Z}_{35}$ is $mathbb{Z}_{35}^#=
left{ 1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34 right}$.

asked Nov 29 '18 at 17:55

vladr10

$C_2 times C_2 times C_6$

$C_2 times C_{12}$

$C_{24}$

The group of units of $mathbb{Z}_{35}$ is $mathbb{Z}_{35}^#=
left{ 1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34 right}$.

group-theory finite-groups

asked Nov 29 '18 at 17:55

vladr10

asked Nov 29 '18 at 17:55

vladr10

asked Nov 29 '18 at 17:55

vladr10

asked Nov 29 '18 at 17:55

vladr10

asked Nov 29 '18 at 17:55

vladr10

closed as off-topic by Derek Holt, Jyrki Lahtonen, DRF, amWhy, ancientmathematician Dec 5 '18 at 17:48

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, DRF, amWhy, ancientmathematician

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

closed as off-topic by Derek Holt, Jyrki Lahtonen, DRF, amWhy, ancientmathematician Dec 5 '18 at 17:48

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, DRF, amWhy, ancientmathematician

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

1

$begingroup$
what have you tried ? Why makes the three possibilities you gave non-isomorphic ?
$endgroup$
– user120527
Nov 29 '18 at 18:03

add a comment |

1

$begingroup$
what have you tried ? Why makes the three possibilities you gave non-isomorphic ?
$endgroup$
– user120527
Nov 29 '18 at 18:03

what have you tried ? Why makes the three possibilities you gave non-isomorphic ?

– user120527
Nov 29 '18 at 18:03

add a comment |

2 Answers
2

active

oldest

votes

Hint: $mathbb{Z}_{35} cong mathbb{Z}_{5} times mathbb{Z}_{7}$, by the Chinese remainder theorem.

edited Nov 30 '18 at 10:10

answered Nov 29 '18 at 17:57

lhf

165k10171395

add a comment |

Chinese Remainder Theorem, so $C_4 times C_6$. Then as pointed out below, you can write as $C_2 times C_3 times C_4 $ and thence as $C_2 times C_{12}$, clumsy though it is.

edited Nov 29 '18 at 18:00

answered Nov 29 '18 at 17:58

Richard Martin

1,61318

$begingroup$
With generators 17 and -1
$endgroup$
– C Monsour
Nov 29 '18 at 18:22

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Hint: $mathbb{Z}_{35} cong mathbb{Z}_{5} times mathbb{Z}_{7}$, by the Chinese remainder theorem.

edited Nov 30 '18 at 10:10

answered Nov 29 '18 at 17:57

lhf

165k10171395

add a comment |

Hint: $mathbb{Z}_{35} cong mathbb{Z}_{5} times mathbb{Z}_{7}$, by the Chinese remainder theorem.

edited Nov 30 '18 at 10:10

answered Nov 29 '18 at 17:57

lhf

165k10171395

add a comment |

Hint: $mathbb{Z}_{35} cong mathbb{Z}_{5} times mathbb{Z}_{7}$, by the Chinese remainder theorem.

edited Nov 30 '18 at 10:10

answered Nov 29 '18 at 17:57

lhf

165k10171395

Hint: $mathbb{Z}_{35} cong mathbb{Z}_{5} times mathbb{Z}_{7}$, by the Chinese remainder theorem.

edited Nov 30 '18 at 10:10

answered Nov 29 '18 at 17:57

lhf

165k10171395

edited Nov 30 '18 at 10:10

answered Nov 29 '18 at 17:57

lhf

165k10171395

answered Nov 29 '18 at 17:57

lhf

165k10171395

answered Nov 29 '18 at 17:57

lhf

165k10171395

add a comment |

Chinese Remainder Theorem, so $C_4 times C_6$. Then as pointed out below, you can write as $C_2 times C_3 times C_4 $ and thence as $C_2 times C_{12}$, clumsy though it is.

edited Nov 29 '18 at 18:00

answered Nov 29 '18 at 17:58

Richard Martin

1,61318

$begingroup$
With generators 17 and -1
$endgroup$
– C Monsour
Nov 29 '18 at 18:22

add a comment |

Chinese Remainder Theorem, so $C_4 times C_6$. Then as pointed out below, you can write as $C_2 times C_3 times C_4 $ and thence as $C_2 times C_{12}$, clumsy though it is.

edited Nov 29 '18 at 18:00

answered Nov 29 '18 at 17:58

Richard Martin

1,61318

$begingroup$
With generators 17 and -1
$endgroup$
– C Monsour
Nov 29 '18 at 18:22

add a comment |

Chinese Remainder Theorem, so $C_4 times C_6$. Then as pointed out below, you can write as $C_2 times C_3 times C_4 $ and thence as $C_2 times C_{12}$, clumsy though it is.

edited Nov 29 '18 at 18:00

answered Nov 29 '18 at 17:58

Richard Martin

1,61318

Chinese Remainder Theorem, so $C_4 times C_6$. Then as pointed out below, you can write as $C_2 times C_3 times C_4 $ and thence as $C_2 times C_{12}$, clumsy though it is.

edited Nov 29 '18 at 18:00

answered Nov 29 '18 at 17:58

Richard Martin

1,61318

edited Nov 29 '18 at 18:00

answered Nov 29 '18 at 17:58

Richard Martin

1,61318

answered Nov 29 '18 at 17:58

Richard Martin

1,61318

answered Nov 29 '18 at 17:58

Richard Martin

1,61318

$begingroup$
With generators 17 and -1
$endgroup$
– C Monsour
Nov 29 '18 at 18:22

add a comment |

$begingroup$
With generators 17 and -1
$endgroup$
– C Monsour
Nov 29 '18 at 18:22

With generators 17 and -1

– C Monsour
Nov 29 '18 at 18:22

add a comment |

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