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Flaw in my classification of groups of order 2015
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In an attempt to classify groups of order $2015 = 5 cdot 13 cdot 31$, I deduced that only $mathbbZ/2015 mathbbZ$ was the only such group. I then checked with some sources that informed me that there are two groups of order 2015 (up to isomorphism) so I was wondering what the flaw in my proof is. I approached the problem as follows:
Let $G$ be a group of order 2015. It is easy to apply Sylow's theorems to show that the Sylow-13 and the Sylow-31 subgroups are normal in $G$; so let $P_13$ and $P_31$ denote the Sylow-13 and Sylow-31 subgroups, respectively. Also let $P_5$ denote a Sylow-5 subgroup.
Since $P_31$ is normal, $P_31P_5$ is a subgroup of order $5 cdot 31$ in $G$. Now since $P_13$ is normal, we have that $P_13 (P_31 P_5)$ is a subgroup of order $13 cdot 31 cdot 5$ in $G$; hence $G = P_13 (P_31 P_5)$. Thus since $P_13 cap (P_31 P_5) = e$, $P_13$ is normal, and $G = P_13 (P_31 P_5)$, we have that $G = P_13 rtimes_theta P_31P_5$. Now since $operatornameAut(P_13) cong mathbbZ /12 mathbbZ $ and $|P_31 P_5|$ is relatively prime to 12, $theta : P_31 P_5 rightarrow operatornameAut(P_13)$ must be trivial. Hence $G = P_13 times P_31 P_5$.
Now I assumed since $P_5$ and $P_31$ are both cyclic there product would be as well, but I am guessing this is where the flaw in my proof is. This is further supported by the fact that there is two groups of order 155. How can I see that there are two possibilities for $P_5P_31$ and conclude that there are only two groups of order 2015?
group-theory finite-groups sylow-theory
asked Jul 29 at 7:46
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1,8551720
add a comment |Â
up vote
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In an attempt to classify groups of order $2015 = 5 cdot 13 cdot 31$, I deduced that only $mathbbZ/2015 mathbbZ$ was the only such group. I then checked with some sources that informed me that there are two groups of order 2015 (up to isomorphism) so I was wondering what the flaw in my proof is. I approached the problem as follows:
Let $G$ be a group of order 2015. It is easy to apply Sylow's theorems to show that the Sylow-13 and the Sylow-31 subgroups are normal in $G$; so let $P_13$ and $P_31$ denote the Sylow-13 and Sylow-31 subgroups, respectively. Also let $P_5$ denote a Sylow-5 subgroup.
Since $P_31$ is normal, $P_31P_5$ is a subgroup of order $5 cdot 31$ in $G$. Now since $P_13$ is normal, we have that $P_13 (P_31 P_5)$ is a subgroup of order $13 cdot 31 cdot 5$ in $G$; hence $G = P_13 (P_31 P_5)$. Thus since $P_13 cap (P_31 P_5) = e$, $P_13$ is normal, and $G = P_13 (P_31 P_5)$, we have that $G = P_13 rtimes_theta P_31P_5$. Now since $operatornameAut(P_13) cong mathbbZ /12 mathbbZ $ and $|P_31 P_5|$ is relatively prime to 12, $theta : P_31 P_5 rightarrow operatornameAut(P_13)$ must be trivial. Hence $G = P_13 times P_31 P_5$.
Now I assumed since $P_5$ and $P_31$ are both cyclic there product would be as well, but I am guessing this is where the flaw in my proof is. This is further supported by the fact that there is two groups of order 155. How can I see that there are two possibilities for $P_5P_31$ and conclude that there are only two groups of order 2015?
group-theory finite-groups sylow-theory
asked Jul 29 at 7:46
Oiler
1,8551720
See this question. There are $2$ groups, see OEIS.
– Dietrich Burde
Jul 29 at 7:56
@DietrichBurde I did see that question, but was unsure with their analysis and wanted to discover the flaw in my proof. And while OEIS is great, I wanted something more constructive as well as something I could recreate in the absence of outside sources.
– Oiler
Jul 29 at 8:02
The answer to the "duplicate" question is at this site (so "inside", if you want); compare the argument with yours.
– Dietrich Burde
Jul 29 at 9:34
add a comment |Â
up vote
5
down vote
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up vote
5
down vote
favorite
In an attempt to classify groups of order $2015 = 5 cdot 13 cdot 31$, I deduced that only $mathbbZ/2015 mathbbZ$ was the only such group. I then checked with some sources that informed me that there are two groups of order 2015 (up to isomorphism) so I was wondering what the flaw in my proof is. I approached the problem as follows:
Let $G$ be a group of order 2015. It is easy to apply Sylow's theorems to show that the Sylow-13 and the Sylow-31 subgroups are normal in $G$; so let $P_13$ and $P_31$ denote the Sylow-13 and Sylow-31 subgroups, respectively. Also let $P_5$ denote a Sylow-5 subgroup.
Since $P_31$ is normal, $P_31P_5$ is a subgroup of order $5 cdot 31$ in $G$. Now since $P_13$ is normal, we have that $P_13 (P_31 P_5)$ is a subgroup of order $13 cdot 31 cdot 5$ in $G$; hence $G = P_13 (P_31 P_5)$. Thus since $P_13 cap (P_31 P_5) = e$, $P_13$ is normal, and $G = P_13 (P_31 P_5)$, we have that $G = P_13 rtimes_theta P_31P_5$. Now since $operatornameAut(P_13) cong mathbbZ /12 mathbbZ $ and $|P_31 P_5|$ is relatively prime to 12, $theta : P_31 P_5 rightarrow operatornameAut(P_13)$ must be trivial. Hence $G = P_13 times P_31 P_5$.
Now I assumed since $P_5$ and $P_31$ are both cyclic there product would be as well, but I am guessing this is where the flaw in my proof is. This is further supported by the fact that there is two groups of order 155. How can I see that there are two possibilities for $P_5P_31$ and conclude that there are only two groups of order 2015?
group-theory finite-groups sylow-theory
asked Jul 29 at 7:46
Oiler
1,8551720
In an attempt to classify groups of order $2015 = 5 cdot 13 cdot 31$, I deduced that only $mathbbZ/2015 mathbbZ$ was the only such group. I then checked with some sources that informed me that there are two groups of order 2015 (up to isomorphism) so I was wondering what the flaw in my proof is. I approached the problem as follows:
Let $G$ be a group of order 2015. It is easy to apply Sylow's theorems to show that the Sylow-13 and the Sylow-31 subgroups are normal in $G$; so let $P_13$ and $P_31$ denote the Sylow-13 and Sylow-31 subgroups, respectively. Also let $P_5$ denote a Sylow-5 subgroup.
Since $P_31$ is normal, $P_31P_5$ is a subgroup of order $5 cdot 31$ in $G$. Now since $P_13$ is normal, we have that $P_13 (P_31 P_5)$ is a subgroup of order $13 cdot 31 cdot 5$ in $G$; hence $G = P_13 (P_31 P_5)$. Thus since $P_13 cap (P_31 P_5) = e$, $P_13$ is normal, and $G = P_13 (P_31 P_5)$, we have that $G = P_13 rtimes_theta P_31P_5$. Now since $operatornameAut(P_13) cong mathbbZ /12 mathbbZ $ and $|P_31 P_5|$ is relatively prime to 12, $theta : P_31 P_5 rightarrow operatornameAut(P_13)$ must be trivial. Hence $G = P_13 times P_31 P_5$.
Now I assumed since $P_5$ and $P_31$ are both cyclic there product would be as well, but I am guessing this is where the flaw in my proof is. This is further supported by the fact that there is two groups of order 155. How can I see that there are two possibilities for $P_5P_31$ and conclude that there are only two groups of order 2015?
group-theory finite-groups sylow-theory
asked Jul 29 at 7:46
Oiler
1,8551720
edited Jul 29 at 7:51
asked Jul 29 at 7:46
Oiler
1,8551720
asked Jul 29 at 7:46
Oiler
1,8551720
asked Jul 29 at 7:46
Oiler
1,8551720
1,8551720
See this question. There are $2$ groups, see OEIS.
– Dietrich Burde
Jul 29 at 7:56
@DietrichBurde I did see that question, but was unsure with their analysis and wanted to discover the flaw in my proof. And while OEIS is great, I wanted something more constructive as well as something I could recreate in the absence of outside sources.
– Oiler
Jul 29 at 8:02
The answer to the "duplicate" question is at this site (so "inside", if you want); compare the argument with yours.
– Dietrich Burde
Jul 29 at 9:34
add a comment |Â
See this question. There are $2$ groups, see OEIS.
– Dietrich Burde
Jul 29 at 7:56
@DietrichBurde I did see that question, but was unsure with their analysis and wanted to discover the flaw in my proof. And while OEIS is great, I wanted something more constructive as well as something I could recreate in the absence of outside sources.
– Oiler
Jul 29 at 8:02
The answer to the "duplicate" question is at this site (so "inside", if you want); compare the argument with yours.
– Dietrich Burde
Jul 29 at 9:34
See this question. There are $2$ groups, see OEIS.
– Dietrich Burde
Jul 29 at 7:56
See this question. There are $2$ groups, see OEIS.
– Dietrich Burde
Jul 29 at 7:56
@DietrichBurde I did see that question, but was unsure with their analysis and wanted to discover the flaw in my proof. And while OEIS is great, I wanted something more constructive as well as something I could recreate in the absence of outside sources.
– Oiler
Jul 29 at 8:02
@DietrichBurde I did see that question, but was unsure with their analysis and wanted to discover the flaw in my proof. And while OEIS is great, I wanted something more constructive as well as something I could recreate in the absence of outside sources.
– Oiler
Jul 29 at 8:02
The answer to the "duplicate" question is at this site (so "inside", if you want); compare the argument with yours.
– Dietrich Burde
Jul 29 at 9:34
The answer to the "duplicate" question is at this site (so "inside", if you want); compare the argument with yours.
– Dietrich Burde
Jul 29 at 9:34
add a comment |Â
1 Answer
1
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up vote
3
down vote
I believe your argument is correct, except for the flaw that you detected yourself.
Let $H$ be a group of order 155. Then there is only one Sylow 31-group $Q_31$ which is hence normal. If $Q_5$ is a normal Sylow 5-group, we have again $H = Q_5 Q_31$, and in fact we have $H = Q_31 rtimes_theta Q_5$ for some action $theta$ of $Q_5$ on $Q_31$. Now, as opposed to your situation above, we have that $5$ does divide $30$, where $operatornameAut(Q_31) cong mathbb Z/30mathbb Z$. Thus, we have to find the morphisms $mathbb Z / 5 mathbb Z to mathbb Z / 30 mathbb Z$. They are either trivial or map $1$ to any element of order $5$, and are determined by that.
If the morphism is trivial, we end up with the direct product and an abelian group. Otherwise, let $theta, chi$ be two actions, and we claim that the resulting groups are isomorphic. Indeed, note that the elements of $mathbb Z / 30 mathbb Z$ of order $5$ are $6,12,18,24$ and form the unique Sylow $5$-group in $mathbb Z / 30 mathbb Z$. By isomorphically relabeling the elements of $Q_31$ except the identity (applying $chi$ and $theta$ to $Q_31$ and writing the results on top of each other and reading e.g. top-down), we can thus transform any such element to another. This procedure yields the desired group isomorphism of semi-direct products.
answered Jul 29 at 8:36
AlgebraicsAnonymous
66611
1
How are you getting that 5,10,15,20,25 have order 5 in Z_30? Do you mean 6, 12, 18, 24?
– JohnC
Aug 1 at 16:18
There are only 15/5 kinds of people: Those who can divide by 5 and those who can't. @JohnC
– AlgebraicsAnonymous
Aug 1 at 20:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I believe your argument is correct, except for the flaw that you detected yourself.
Let $H$ be a group of order 155. Then there is only one Sylow 31-group $Q_31$ which is hence normal. If $Q_5$ is a normal Sylow 5-group, we have again $H = Q_5 Q_31$, and in fact we have $H = Q_31 rtimes_theta Q_5$ for some action $theta$ of $Q_5$ on $Q_31$. Now, as opposed to your situation above, we have that $5$ does divide $30$, where $operatornameAut(Q_31) cong mathbb Z/30mathbb Z$. Thus, we have to find the morphisms $mathbb Z / 5 mathbb Z to mathbb Z / 30 mathbb Z$. They are either trivial or map $1$ to any element of order $5$, and are determined by that.
If the morphism is trivial, we end up with the direct product and an abelian group. Otherwise, let $theta, chi$ be two actions, and we claim that the resulting groups are isomorphic. Indeed, note that the elements of $mathbb Z / 30 mathbb Z$ of order $5$ are $6,12,18,24$ and form the unique Sylow $5$-group in $mathbb Z / 30 mathbb Z$. By isomorphically relabeling the elements of $Q_31$ except the identity (applying $chi$ and $theta$ to $Q_31$ and writing the results on top of each other and reading e.g. top-down), we can thus transform any such element to another. This procedure yields the desired group isomorphism of semi-direct products.
answered Jul 29 at 8:36
AlgebraicsAnonymous
66611
1
How are you getting that 5,10,15,20,25 have order 5 in Z_30? Do you mean 6, 12, 18, 24?
– JohnC
Aug 1 at 16:18
There are only 15/5 kinds of people: Those who can divide by 5 and those who can't. @JohnC
– AlgebraicsAnonymous
Aug 1 at 20:03
add a comment |Â
up vote
3
down vote
I believe your argument is correct, except for the flaw that you detected yourself.
Let $H$ be a group of order 155. Then there is only one Sylow 31-group $Q_31$ which is hence normal. If $Q_5$ is a normal Sylow 5-group, we have again $H = Q_5 Q_31$, and in fact we have $H = Q_31 rtimes_theta Q_5$ for some action $theta$ of $Q_5$ on $Q_31$. Now, as opposed to your situation above, we have that $5$ does divide $30$, where $operatornameAut(Q_31) cong mathbb Z/30mathbb Z$. Thus, we have to find the morphisms $mathbb Z / 5 mathbb Z to mathbb Z / 30 mathbb Z$. They are either trivial or map $1$ to any element of order $5$, and are determined by that.
If the morphism is trivial, we end up with the direct product and an abelian group. Otherwise, let $theta, chi$ be two actions, and we claim that the resulting groups are isomorphic. Indeed, note that the elements of $mathbb Z / 30 mathbb Z$ of order $5$ are $6,12,18,24$ and form the unique Sylow $5$-group in $mathbb Z / 30 mathbb Z$. By isomorphically relabeling the elements of $Q_31$ except the identity (applying $chi$ and $theta$ to $Q_31$ and writing the results on top of each other and reading e.g. top-down), we can thus transform any such element to another. This procedure yields the desired group isomorphism of semi-direct products.
answered Jul 29 at 8:36
AlgebraicsAnonymous
66611
1
How are you getting that 5,10,15,20,25 have order 5 in Z_30? Do you mean 6, 12, 18, 24?
– JohnC
Aug 1 at 16:18
There are only 15/5 kinds of people: Those who can divide by 5 and those who can't. @JohnC
– AlgebraicsAnonymous
Aug 1 at 20:03
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I believe your argument is correct, except for the flaw that you detected yourself.
Let $H$ be a group of order 155. Then there is only one Sylow 31-group $Q_31$ which is hence normal. If $Q_5$ is a normal Sylow 5-group, we have again $H = Q_5 Q_31$, and in fact we have $H = Q_31 rtimes_theta Q_5$ for some action $theta$ of $Q_5$ on $Q_31$. Now, as opposed to your situation above, we have that $5$ does divide $30$, where $operatornameAut(Q_31) cong mathbb Z/30mathbb Z$. Thus, we have to find the morphisms $mathbb Z / 5 mathbb Z to mathbb Z / 30 mathbb Z$. They are either trivial or map $1$ to any element of order $5$, and are determined by that.
If the morphism is trivial, we end up with the direct product and an abelian group. Otherwise, let $theta, chi$ be two actions, and we claim that the resulting groups are isomorphic. Indeed, note that the elements of $mathbb Z / 30 mathbb Z$ of order $5$ are $6,12,18,24$ and form the unique Sylow $5$-group in $mathbb Z / 30 mathbb Z$. By isomorphically relabeling the elements of $Q_31$ except the identity (applying $chi$ and $theta$ to $Q_31$ and writing the results on top of each other and reading e.g. top-down), we can thus transform any such element to another. This procedure yields the desired group isomorphism of semi-direct products.
answered Jul 29 at 8:36
AlgebraicsAnonymous
66611
I believe your argument is correct, except for the flaw that you detected yourself.
Let $H$ be a group of order 155. Then there is only one Sylow 31-group $Q_31$ which is hence normal. If $Q_5$ is a normal Sylow 5-group, we have again $H = Q_5 Q_31$, and in fact we have $H = Q_31 rtimes_theta Q_5$ for some action $theta$ of $Q_5$ on $Q_31$. Now, as opposed to your situation above, we have that $5$ does divide $30$, where $operatornameAut(Q_31) cong mathbb Z/30mathbb Z$. Thus, we have to find the morphisms $mathbb Z / 5 mathbb Z to mathbb Z / 30 mathbb Z$. They are either trivial or map $1$ to any element of order $5$, and are determined by that.
If the morphism is trivial, we end up with the direct product and an abelian group. Otherwise, let $theta, chi$ be two actions, and we claim that the resulting groups are isomorphic. Indeed, note that the elements of $mathbb Z / 30 mathbb Z$ of order $5$ are $6,12,18,24$ and form the unique Sylow $5$-group in $mathbb Z / 30 mathbb Z$. By isomorphically relabeling the elements of $Q_31$ except the identity (applying $chi$ and $theta$ to $Q_31$ and writing the results on top of each other and reading e.g. top-down), we can thus transform any such element to another. This procedure yields the desired group isomorphism of semi-direct products.
answered Jul 29 at 8:36
AlgebraicsAnonymous
66611
edited Aug 1 at 20:02
answered Jul 29 at 8:36
AlgebraicsAnonymous
66611
answered Jul 29 at 8:36
AlgebraicsAnonymous
66611
answered Jul 29 at 8:36
AlgebraicsAnonymous
66611
66611
1
How are you getting that 5,10,15,20,25 have order 5 in Z_30? Do you mean 6, 12, 18, 24?
– JohnC
Aug 1 at 16:18
There are only 15/5 kinds of people: Those who can divide by 5 and those who can't. @JohnC
– AlgebraicsAnonymous
Aug 1 at 20:03
add a comment |Â
1
How are you getting that 5,10,15,20,25 have order 5 in Z_30? Do you mean 6, 12, 18, 24?
– JohnC
Aug 1 at 16:18
There are only 15/5 kinds of people: Those who can divide by 5 and those who can't. @JohnC
– AlgebraicsAnonymous
Aug 1 at 20:03
1
1
How are you getting that 5,10,15,20,25 have order 5 in Z_30? Do you mean 6, 12, 18, 24?
– JohnC
Aug 1 at 16:18
How are you getting that 5,10,15,20,25 have order 5 in Z_30? Do you mean 6, 12, 18, 24?
– JohnC
Aug 1 at 16:18
There are only 15/5 kinds of people: Those who can divide by 5 and those who can't. @JohnC
– AlgebraicsAnonymous
Aug 1 at 20:03
There are only 15/5 kinds of people: Those who can divide by 5 and those who can't. @JohnC
– AlgebraicsAnonymous
Aug 1 at 20:03
add a comment |Â
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See this question. There are $2$ groups, see OEIS.
– Dietrich Burde
Jul 29 at 7:56
@DietrichBurde I did see that question, but was unsure with their analysis and wanted to discover the flaw in my proof. And while OEIS is great, I wanted something more constructive as well as something I could recreate in the absence of outside sources.
– Oiler
Jul 29 at 8:02
The answer to the "duplicate" question is at this site (so "inside", if you want); compare the argument with yours.
– Dietrich Burde
Jul 29 at 9:34