Find two finite groups $G_1$ and $G_2$ such as:

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Find two finite groups $G_1$ and $G_2$ such as:



$1)$ $|G_1|=|G_2|$



$2)$ for all prime $p$ every $p$-sylow subgroup of $G_1$ isomorpic to $p$-sylow subgroup of $G_2$.



but $G_1notcong G_2$.



hello I tried to find an example but I couldn't find one please help me.
I look on all the groups from order 6 (I choose this order randomly) but I don't know how to prove that every p-sylow subgroup of $G_1$ isomorpic to every p-sylow subgroup of $G_2$ for all primes p.







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  • Maybe try to add what you have already tried.
    – zzuussee
    Aug 3 at 11:46






  • 1




    Have you seen semi-direct product? If so $C_21$ and $C_3ltimes C_7$ would do.
    – daruma
    Aug 3 at 11:49











  • The first version of your question was not very detailed, but your edit is already better. You can take a look at this page to find some advice about what you should put in your question : math.meta.stackexchange.com/questions/9959/…
    – Arnaud D.
    Aug 3 at 11:55














up vote
1
down vote

favorite












Find two finite groups $G_1$ and $G_2$ such as:



$1)$ $|G_1|=|G_2|$



$2)$ for all prime $p$ every $p$-sylow subgroup of $G_1$ isomorpic to $p$-sylow subgroup of $G_2$.



but $G_1notcong G_2$.



hello I tried to find an example but I couldn't find one please help me.
I look on all the groups from order 6 (I choose this order randomly) but I don't know how to prove that every p-sylow subgroup of $G_1$ isomorpic to every p-sylow subgroup of $G_2$ for all primes p.







share|cite|improve this question





















  • Maybe try to add what you have already tried.
    – zzuussee
    Aug 3 at 11:46






  • 1




    Have you seen semi-direct product? If so $C_21$ and $C_3ltimes C_7$ would do.
    – daruma
    Aug 3 at 11:49











  • The first version of your question was not very detailed, but your edit is already better. You can take a look at this page to find some advice about what you should put in your question : math.meta.stackexchange.com/questions/9959/…
    – Arnaud D.
    Aug 3 at 11:55












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Find two finite groups $G_1$ and $G_2$ such as:



$1)$ $|G_1|=|G_2|$



$2)$ for all prime $p$ every $p$-sylow subgroup of $G_1$ isomorpic to $p$-sylow subgroup of $G_2$.



but $G_1notcong G_2$.



hello I tried to find an example but I couldn't find one please help me.
I look on all the groups from order 6 (I choose this order randomly) but I don't know how to prove that every p-sylow subgroup of $G_1$ isomorpic to every p-sylow subgroup of $G_2$ for all primes p.







share|cite|improve this question













Find two finite groups $G_1$ and $G_2$ such as:



$1)$ $|G_1|=|G_2|$



$2)$ for all prime $p$ every $p$-sylow subgroup of $G_1$ isomorpic to $p$-sylow subgroup of $G_2$.



but $G_1notcong G_2$.



hello I tried to find an example but I couldn't find one please help me.
I look on all the groups from order 6 (I choose this order randomly) but I don't know how to prove that every p-sylow subgroup of $G_1$ isomorpic to every p-sylow subgroup of $G_2$ for all primes p.









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 11:52
























asked Aug 3 at 11:44









Rimon

33




33











  • Maybe try to add what you have already tried.
    – zzuussee
    Aug 3 at 11:46






  • 1




    Have you seen semi-direct product? If so $C_21$ and $C_3ltimes C_7$ would do.
    – daruma
    Aug 3 at 11:49











  • The first version of your question was not very detailed, but your edit is already better. You can take a look at this page to find some advice about what you should put in your question : math.meta.stackexchange.com/questions/9959/…
    – Arnaud D.
    Aug 3 at 11:55
















  • Maybe try to add what you have already tried.
    – zzuussee
    Aug 3 at 11:46






  • 1




    Have you seen semi-direct product? If so $C_21$ and $C_3ltimes C_7$ would do.
    – daruma
    Aug 3 at 11:49











  • The first version of your question was not very detailed, but your edit is already better. You can take a look at this page to find some advice about what you should put in your question : math.meta.stackexchange.com/questions/9959/…
    – Arnaud D.
    Aug 3 at 11:55















Maybe try to add what you have already tried.
– zzuussee
Aug 3 at 11:46




Maybe try to add what you have already tried.
– zzuussee
Aug 3 at 11:46




1




1




Have you seen semi-direct product? If so $C_21$ and $C_3ltimes C_7$ would do.
– daruma
Aug 3 at 11:49





Have you seen semi-direct product? If so $C_21$ and $C_3ltimes C_7$ would do.
– daruma
Aug 3 at 11:49













The first version of your question was not very detailed, but your edit is already better. You can take a look at this page to find some advice about what you should put in your question : math.meta.stackexchange.com/questions/9959/…
– Arnaud D.
Aug 3 at 11:55




The first version of your question was not very detailed, but your edit is already better. You can take a look at this page to find some advice about what you should put in your question : math.meta.stackexchange.com/questions/9959/…
– Arnaud D.
Aug 3 at 11:55










1 Answer
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up vote
4
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If $p, q$ are distinct primes then every two groups of order $pq$ will have isomorphic Sylow subgroups (being cyclic subgroups of order $p$ and $q$). So all you have to do is to find two nonisomorphic subgroups of order $pq$, e.g. $mathbbZ_6$ and $S_3$.






share|cite|improve this answer





















  • thank you :) ,I am sorry but how do I prove that every p-sylow subgroups ane isomorpic?
    – Rimon
    Aug 3 at 11:58






  • 1




    @Rimon Every two cyclic groups of the same order are isomorphic. And every group of order $pq$ has both: subgroups of order $p$ and of order $q$ (Cauchy's theorem). These subgroups are necessary Sylow. That's everything you need.
    – freakish
    Aug 3 at 12:01











  • thank you :) I get it now
    – Rimon
    Aug 3 at 12:30










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1 Answer
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1 Answer
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active

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up vote
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If $p, q$ are distinct primes then every two groups of order $pq$ will have isomorphic Sylow subgroups (being cyclic subgroups of order $p$ and $q$). So all you have to do is to find two nonisomorphic subgroups of order $pq$, e.g. $mathbbZ_6$ and $S_3$.






share|cite|improve this answer





















  • thank you :) ,I am sorry but how do I prove that every p-sylow subgroups ane isomorpic?
    – Rimon
    Aug 3 at 11:58






  • 1




    @Rimon Every two cyclic groups of the same order are isomorphic. And every group of order $pq$ has both: subgroups of order $p$ and of order $q$ (Cauchy's theorem). These subgroups are necessary Sylow. That's everything you need.
    – freakish
    Aug 3 at 12:01











  • thank you :) I get it now
    – Rimon
    Aug 3 at 12:30














up vote
4
down vote













If $p, q$ are distinct primes then every two groups of order $pq$ will have isomorphic Sylow subgroups (being cyclic subgroups of order $p$ and $q$). So all you have to do is to find two nonisomorphic subgroups of order $pq$, e.g. $mathbbZ_6$ and $S_3$.






share|cite|improve this answer





















  • thank you :) ,I am sorry but how do I prove that every p-sylow subgroups ane isomorpic?
    – Rimon
    Aug 3 at 11:58






  • 1




    @Rimon Every two cyclic groups of the same order are isomorphic. And every group of order $pq$ has both: subgroups of order $p$ and of order $q$ (Cauchy's theorem). These subgroups are necessary Sylow. That's everything you need.
    – freakish
    Aug 3 at 12:01











  • thank you :) I get it now
    – Rimon
    Aug 3 at 12:30












up vote
4
down vote










up vote
4
down vote









If $p, q$ are distinct primes then every two groups of order $pq$ will have isomorphic Sylow subgroups (being cyclic subgroups of order $p$ and $q$). So all you have to do is to find two nonisomorphic subgroups of order $pq$, e.g. $mathbbZ_6$ and $S_3$.






share|cite|improve this answer













If $p, q$ are distinct primes then every two groups of order $pq$ will have isomorphic Sylow subgroups (being cyclic subgroups of order $p$ and $q$). So all you have to do is to find two nonisomorphic subgroups of order $pq$, e.g. $mathbbZ_6$ and $S_3$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 3 at 11:50









freakish

8,4371524




8,4371524











  • thank you :) ,I am sorry but how do I prove that every p-sylow subgroups ane isomorpic?
    – Rimon
    Aug 3 at 11:58






  • 1




    @Rimon Every two cyclic groups of the same order are isomorphic. And every group of order $pq$ has both: subgroups of order $p$ and of order $q$ (Cauchy's theorem). These subgroups are necessary Sylow. That's everything you need.
    – freakish
    Aug 3 at 12:01











  • thank you :) I get it now
    – Rimon
    Aug 3 at 12:30
















  • thank you :) ,I am sorry but how do I prove that every p-sylow subgroups ane isomorpic?
    – Rimon
    Aug 3 at 11:58






  • 1




    @Rimon Every two cyclic groups of the same order are isomorphic. And every group of order $pq$ has both: subgroups of order $p$ and of order $q$ (Cauchy's theorem). These subgroups are necessary Sylow. That's everything you need.
    – freakish
    Aug 3 at 12:01











  • thank you :) I get it now
    – Rimon
    Aug 3 at 12:30















thank you :) ,I am sorry but how do I prove that every p-sylow subgroups ane isomorpic?
– Rimon
Aug 3 at 11:58




thank you :) ,I am sorry but how do I prove that every p-sylow subgroups ane isomorpic?
– Rimon
Aug 3 at 11:58




1




1




@Rimon Every two cyclic groups of the same order are isomorphic. And every group of order $pq$ has both: subgroups of order $p$ and of order $q$ (Cauchy's theorem). These subgroups are necessary Sylow. That's everything you need.
– freakish
Aug 3 at 12:01





@Rimon Every two cyclic groups of the same order are isomorphic. And every group of order $pq$ has both: subgroups of order $p$ and of order $q$ (Cauchy's theorem). These subgroups are necessary Sylow. That's everything you need.
– freakish
Aug 3 at 12:01













thank you :) I get it now
– Rimon
Aug 3 at 12:30




thank you :) I get it now
– Rimon
Aug 3 at 12:30












 

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