P-sylow subgroups of group of order $340$

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I have to prove that a group of order $340$ has a cyclic normal subgroup of order $85$. Because $$340=17times5times2^2$$ and due to the third Sylow theorem that there is only one $17$-Sylow subgroup and one $5$-Sylow subgroup, so they are normal. So the group of order $85$ is also normal.

The problem is that, I know that, it has to be isomorphic to $BbbZ_85$ or $BbbZ_17 times BbbZ_5$ and I don't know how to prove that it's the first one. Any advice? Thanks

group-theory sylow-theory

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edited Jul 29 at 15:51

Michael Hardy

204k23185461

asked Jul 29 at 15:05

J. González

132

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Note that a group of order $85$ must be cyclic because of $5nmid 17-1$
  – Peter
  Jul 29 at 15:10
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  In fact, $mathbb Z_85$ is isomorphic to $mathbb Z_17times mathbb Z_5$
  – Peter
  Jul 29 at 15:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Oh youre right, thank you so much
  – J. González
  Jul 29 at 15:25
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

I have to prove that a group of order $340$ has a cyclic normal subgroup of order $85$. Because $$340=17times5times2^2$$ and due to the third Sylow theorem that there is only one $17$-Sylow subgroup and one $5$-Sylow subgroup, so they are normal. So the group of order $85$ is also normal.

The problem is that, I know that, it has to be isomorphic to $BbbZ_85$ or $BbbZ_17 times BbbZ_5$ and I don't know how to prove that it's the first one. Any advice? Thanks

group-theory sylow-theory

share|cite|improve this question

edited Jul 29 at 15:51

Michael Hardy

204k23185461

asked Jul 29 at 15:05

J. González

132

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Note that a group of order $85$ must be cyclic because of $5nmid 17-1$
  – Peter
  Jul 29 at 15:10
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  In fact, $mathbb Z_85$ is isomorphic to $mathbb Z_17times mathbb Z_5$
  – Peter
  Jul 29 at 15:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Oh youre right, thank you so much
  – J. González
  Jul 29 at 15:25
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

I have to prove that a group of order $340$ has a cyclic normal subgroup of order $85$. Because $$340=17times5times2^2$$ and due to the third Sylow theorem that there is only one $17$-Sylow subgroup and one $5$-Sylow subgroup, so they are normal. So the group of order $85$ is also normal.

The problem is that, I know that, it has to be isomorphic to $BbbZ_85$ or $BbbZ_17 times BbbZ_5$ and I don't know how to prove that it's the first one. Any advice? Thanks

group-theory sylow-theory

share|cite|improve this question

edited Jul 29 at 15:51

Michael Hardy

204k23185461

asked Jul 29 at 15:05

J. González

132

I have to prove that a group of order $340$ has a cyclic normal subgroup of order $85$. Because $$340=17times5times2^2$$ and due to the third Sylow theorem that there is only one $17$-Sylow subgroup and one $5$-Sylow subgroup, so they are normal. So the group of order $85$ is also normal.

The problem is that, I know that, it has to be isomorphic to $BbbZ_85$ or $BbbZ_17 times BbbZ_5$ and I don't know how to prove that it's the first one. Any advice? Thanks

group-theory sylow-theory

share|cite|improve this question

edited Jul 29 at 15:51

Michael Hardy

204k23185461

asked Jul 29 at 15:05

J. González

132

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 15:51

Michael Hardy

204k23185461

edited Jul 29 at 15:51

Michael Hardy

204k23185461

edited Jul 29 at 15:51

Michael Hardy

204k23185461

204k23185461

asked Jul 29 at 15:05

J. González

132

asked Jul 29 at 15:05

J. González

132

asked Jul 29 at 15:05

J. González

132

132

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Note that a group of order $85$ must be cyclic because of $5nmid 17-1$
  – Peter
  Jul 29 at 15:10
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  In fact, $mathbb Z_85$ is isomorphic to $mathbb Z_17times mathbb Z_5$
  – Peter
  Jul 29 at 15:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Oh youre right, thank you so much
  – J. González
  Jul 29 at 15:25
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Note that a group of order $85$ must be cyclic because of $5nmid 17-1$
  – Peter
  Jul 29 at 15:10
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  In fact, $mathbb Z_85$ is isomorphic to $mathbb Z_17times mathbb Z_5$
  – Peter
  Jul 29 at 15:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Oh youre right, thank you so much
  – J. González
  Jul 29 at 15:25
  

  
  
  
  

3

3

Note that a group of order $85$ must be cyclic because of $5nmid 17-1$
– Peter
Jul 29 at 15:10

Note that a group of order $85$ must be cyclic because of $5nmid 17-1$
– Peter
Jul 29 at 15:10

3

3

In fact, $mathbb Z_85$ is isomorphic to $mathbb Z_17times mathbb Z_5$
– Peter
Jul 29 at 15:13

In fact, $mathbb Z_85$ is isomorphic to $mathbb Z_17times mathbb Z_5$
– Peter
Jul 29 at 15:13

1

1

Oh youre right, thank you so much
– J. González
Jul 29 at 15:25

Oh youre right, thank you so much
– J. González
Jul 29 at 15:25

add a comment | 

1 Answer
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Since $Bbb Z_17$ has an element $h$ of order $17$ and $Bbb Z_5$ has an element $k$ of order $5,$ then $langle h,krangle$ is an element of $Bbb Z_17timesBbb Z_5,$ whose order is $operatornamelcm(17,5)=85.$ Since we readily have $bigllvertBbb Z_17timesBbb Z_5bigrrvert=85,$ then this tells us that $Bbb Z_17timesBbb Z_5$ is a cyclic group of order $85,$ meaning that it's isomorphic to $Bbb Z_85.$

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answered Jul 29 at 16:24

Cameron Buie

83.5k771152

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1 Answer
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1 Answer
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up vote
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accepted

Since $Bbb Z_17$ has an element $h$ of order $17$ and $Bbb Z_5$ has an element $k$ of order $5,$ then $langle h,krangle$ is an element of $Bbb Z_17timesBbb Z_5,$ whose order is $operatornamelcm(17,5)=85.$ Since we readily have $bigllvertBbb Z_17timesBbb Z_5bigrrvert=85,$ then this tells us that $Bbb Z_17timesBbb Z_5$ is a cyclic group of order $85,$ meaning that it's isomorphic to $Bbb Z_85.$

share|cite|improve this answer

answered Jul 29 at 16:24

Cameron Buie

83.5k771152

add a comment | 

up vote
1
down vote



accepted

Since $Bbb Z_17$ has an element $h$ of order $17$ and $Bbb Z_5$ has an element $k$ of order $5,$ then $langle h,krangle$ is an element of $Bbb Z_17timesBbb Z_5,$ whose order is $operatornamelcm(17,5)=85.$ Since we readily have $bigllvertBbb Z_17timesBbb Z_5bigrrvert=85,$ then this tells us that $Bbb Z_17timesBbb Z_5$ is a cyclic group of order $85,$ meaning that it's isomorphic to $Bbb Z_85.$

share|cite|improve this answer

answered Jul 29 at 16:24

Cameron Buie

83.5k771152

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Since $Bbb Z_17$ has an element $h$ of order $17$ and $Bbb Z_5$ has an element $k$ of order $5,$ then $langle h,krangle$ is an element of $Bbb Z_17timesBbb Z_5,$ whose order is $operatornamelcm(17,5)=85.$ Since we readily have $bigllvertBbb Z_17timesBbb Z_5bigrrvert=85,$ then this tells us that $Bbb Z_17timesBbb Z_5$ is a cyclic group of order $85,$ meaning that it's isomorphic to $Bbb Z_85.$

share|cite|improve this answer

answered Jul 29 at 16:24

Cameron Buie

83.5k771152

Since $Bbb Z_17$ has an element $h$ of order $17$ and $Bbb Z_5$ has an element $k$ of order $5,$ then $langle h,krangle$ is an element of $Bbb Z_17timesBbb Z_5,$ whose order is $operatornamelcm(17,5)=85.$ Since we readily have $bigllvertBbb Z_17timesBbb Z_5bigrrvert=85,$ then this tells us that $Bbb Z_17timesBbb Z_5$ is a cyclic group of order $85,$ meaning that it's isomorphic to $Bbb Z_85.$

share|cite|improve this answer

answered Jul 29 at 16:24

Cameron Buie

83.5k771152

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 16:24

Cameron Buie

83.5k771152

answered Jul 29 at 16:24

Cameron Buie

83.5k771152

answered Jul 29 at 16:24

Cameron Buie

83.5k771152

83.5k771152

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