Mixing
Explicit descriptions of groups of order 45
I know that there are two groups of order 45, and obviously one of them (up to isomorphism) is $mathbb{Z}_{45}$. I'm trying to understand explicitly what the structure of the other is like.
By Cauchy's Theorem and Sylow's First Theorem, it has a subgroup of order 3, one of order 5, and one of order 9. What this second group of order 45 doesn't have, unlike $mathbb{Z}_{45}$, is a subgroup of order 15. This rules out an element of order 15, which would generate a cyclic subgroup. I know the group is abelian since we can write it as a product of normal subgroups. Otherwise I'm unsure of what consequences this has, and what the structure of the non-cyclic group of order 45 is.
So basically my question is...how do we obtain an explicit description of the the non-cyclic group of order 45?
group-theory finite-groups
edited Feb 12 '12 at 5:21
Arturo Magidin
261k32584904
asked Feb 12 '12 at 5:19
Alex Petzke
3,97423569
add a comment |
I know that there are two groups of order 45, and obviously one of them (up to isomorphism) is $mathbb{Z}_{45}$. I'm trying to understand explicitly what the structure of the other is like.
By Cauchy's Theorem and Sylow's First Theorem, it has a subgroup of order 3, one of order 5, and one of order 9. What this second group of order 45 doesn't have, unlike $mathbb{Z}_{45}$, is a subgroup of order 15. This rules out an element of order 15, which would generate a cyclic subgroup. I know the group is abelian since we can write it as a product of normal subgroups. Otherwise I'm unsure of what consequences this has, and what the structure of the non-cyclic group of order 45 is.
So basically my question is...how do we obtain an explicit description of the the non-cyclic group of order 45?
group-theory finite-groups
edited Feb 12 '12 at 5:21
Arturo Magidin
261k32584904
asked Feb 12 '12 at 5:19
Alex Petzke
3,97423569
add a comment |
I know that there are two groups of order 45, and obviously one of them (up to isomorphism) is $mathbb{Z}_{45}$. I'm trying to understand explicitly what the structure of the other is like.
By Cauchy's Theorem and Sylow's First Theorem, it has a subgroup of order 3, one of order 5, and one of order 9. What this second group of order 45 doesn't have, unlike $mathbb{Z}_{45}$, is a subgroup of order 15. This rules out an element of order 15, which would generate a cyclic subgroup. I know the group is abelian since we can write it as a product of normal subgroups. Otherwise I'm unsure of what consequences this has, and what the structure of the non-cyclic group of order 45 is.
So basically my question is...how do we obtain an explicit description of the the non-cyclic group of order 45?
group-theory finite-groups
edited Feb 12 '12 at 5:21
Arturo Magidin
261k32584904
asked Feb 12 '12 at 5:19
Alex Petzke
3,97423569
I know that there are two groups of order 45, and obviously one of them (up to isomorphism) is $mathbb{Z}_{45}$. I'm trying to understand explicitly what the structure of the other is like.
By Cauchy's Theorem and Sylow's First Theorem, it has a subgroup of order 3, one of order 5, and one of order 9. What this second group of order 45 doesn't have, unlike $mathbb{Z}_{45}$, is a subgroup of order 15. This rules out an element of order 15, which would generate a cyclic subgroup. I know the group is abelian since we can write it as a product of normal subgroups. Otherwise I'm unsure of what consequences this has, and what the structure of the non-cyclic group of order 45 is.
So basically my question is...how do we obtain an explicit description of the the non-cyclic group of order 45?
group-theory finite-groups
group-theory finite-groups
edited Feb 12 '12 at 5:21
Arturo Magidin
261k32584904
asked Feb 12 '12 at 5:19
Alex Petzke
3,97423569
edited Feb 12 '12 at 5:21
Arturo Magidin
261k32584904
asked Feb 12 '12 at 5:19
Alex Petzke
3,97423569
edited Feb 12 '12 at 5:21
Arturo Magidin
261k32584904
edited Feb 12 '12 at 5:21
Arturo Magidin
261k32584904
edited Feb 12 '12 at 5:21
Arturo Magidin
261k32584904
261k32584904
asked Feb 12 '12 at 5:19
Alex Petzke
3,97423569
asked Feb 12 '12 at 5:19
Alex Petzke
3,97423569
asked Feb 12 '12 at 5:19
Alex Petzke
3,97423569
3,97423569
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
By Sylow's theorem, the number of subgroups of order $5$ must divide $45$, and must be congruent to $1$ modulo $5$; the only possibility is that there is a single group of order $5$. Likewise, the number of subgroups of order $9$ must divide $45$ and be congruent to $1$ modulo $3$; the only possibility is that there is a single subgroup of order $9$.
So the group has a single subgroup of order $5$, which must be isomorphic to $mathbb{Z}_5$ (since that is the only group of order $5$), and a single subgroup of order $9$, which is isomorphic to either $mathbb{Z}_9$ or to $mathbb{Z}_3timesmathbb{Z}_3$ (since the only groups of order $p^2$, with $p$ prime, are the cyclic group of order $p^2$ and a direct product of two copies of the cyclic group of order $p$).
Moreover, both Sylow subgroups are normal; let $N$ be the $5$-Sylow subgroups, and $M$ be the $3$-Sylow subgroup. Then $Ncap M={1}$ since the orders are coprime, and $NM$ is a subgroup (since both $N$ and $M$ are normal). And
$$|NM| = frac{|N|,|M|}{|Ncap M|} = 9times 5 = 45 = |G|,$$
so $G=NM$. Since $N$ and $M$ are normal, $nm=mn$ for every $min M$ and $nin N$. So $Gcong Ntimes Mcong mathbb{Z}_5times M$. There are two possibilities for $M$, giving you the only two possibilities for $G$.
Note that in both cases you get subgroups of order $15$; what you don't get in one of the cases is an element of order $45$.
answered Feb 12 '12 at 5:27
Arturo Magidin
261k32584904
You wrote $3 times 5 = 45$. Lol
– Patrick Da Silva
Feb 12 '12 at 5:31
@Patrick: And you wrote $mathbb{Z}_3timesmathbb{Z}_3mathbb{Z}_5$... Fixed.
– Arturo Magidin
Feb 12 '12 at 5:33
Great answer. You did not assume OP's finding which was that there was only $2$ groups of order $45$, which I think is cool for people who have no idea how OP might have found that.
– Patrick Da Silva
Feb 12 '12 at 5:34
I had nearly all of that in mind. All except the two possibilities for the Sylow 3-subgroup. This idea of decomposing the subgroups I'm sure generalizes very widely, so this is great. Thanks.
– Alex Petzke
Feb 12 '12 at 13:15
add a comment |
If you know that there are two groups of order $45$, the first groups you must think of are abelian groups, i.e. direct product of cyclic groups by the finitely-generated-abelian-groups classification. Now your "building blocks" are $mathbb Z_3$, which you have twice, and $mathbb Z_5$, which you have once. If you try to "put them all together" you get $mathbb Z_{45}$, which we clearly expected. Now other combinations would be $mathbb Z_9 times mathbb Z_5$, $mathbb Z_3 times mathbb Z_3times mathbb Z_5$, and $mathbb Z_{15} times mathbb Z_3$. For the first one, since $g.c.d.(9,5) = 1$, you know that this group will be cyclic, hence isomorphic to $mathbb Z_{45}$.
For the other two, since $g.c.d(3,5) = 1$, $mathbb Z_3 times mathbb Z_5$ is isomorphic to $mathbb Z_{15}$ so we don't have anything new there. But in $mathbb Z_{15} times mathbb Z_3$, every element has order that divides $15$, thus this group cannot be isomorphic to $mathbb Z_{45}$.
Hope that helps,
edited Feb 12 '12 at 5:32
Arturo Magidin
261k32584904
answered Feb 12 '12 at 5:29
Patrick Da Silva
32k353106
Heh you were quicker than me on that one!
– Patrick Da Silva
Feb 12 '12 at 5:33
Yes, knowing that it was abelian I should have followed up with the Fundamental Theorem of Finite Abelian Groups... Another good way of looking at it. Thanks.
– Alex Petzke
Feb 12 '12 at 20:22
add a comment |
It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.
answered Feb 12 '12 at 5:23
Jim Belk
37.4k285150
add a comment |
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3 Answers
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3 Answers
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oldest
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oldest
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By Sylow's theorem, the number of subgroups of order $5$ must divide $45$, and must be congruent to $1$ modulo $5$; the only possibility is that there is a single group of order $5$. Likewise, the number of subgroups of order $9$ must divide $45$ and be congruent to $1$ modulo $3$; the only possibility is that there is a single subgroup of order $9$.
So the group has a single subgroup of order $5$, which must be isomorphic to $mathbb{Z}_5$ (since that is the only group of order $5$), and a single subgroup of order $9$, which is isomorphic to either $mathbb{Z}_9$ or to $mathbb{Z}_3timesmathbb{Z}_3$ (since the only groups of order $p^2$, with $p$ prime, are the cyclic group of order $p^2$ and a direct product of two copies of the cyclic group of order $p$).
Moreover, both Sylow subgroups are normal; let $N$ be the $5$-Sylow subgroups, and $M$ be the $3$-Sylow subgroup. Then $Ncap M={1}$ since the orders are coprime, and $NM$ is a subgroup (since both $N$ and $M$ are normal). And
$$|NM| = frac{|N|,|M|}{|Ncap M|} = 9times 5 = 45 = |G|,$$
so $G=NM$. Since $N$ and $M$ are normal, $nm=mn$ for every $min M$ and $nin N$. So $Gcong Ntimes Mcong mathbb{Z}_5times M$. There are two possibilities for $M$, giving you the only two possibilities for $G$.
Note that in both cases you get subgroups of order $15$; what you don't get in one of the cases is an element of order $45$.
answered Feb 12 '12 at 5:27
Arturo Magidin
261k32584904
You wrote $3 times 5 = 45$. Lol
– Patrick Da Silva
Feb 12 '12 at 5:31
@Patrick: And you wrote $mathbb{Z}_3timesmathbb{Z}_3mathbb{Z}_5$... Fixed.
– Arturo Magidin
Feb 12 '12 at 5:33
Great answer. You did not assume OP's finding which was that there was only $2$ groups of order $45$, which I think is cool for people who have no idea how OP might have found that.
– Patrick Da Silva
Feb 12 '12 at 5:34
I had nearly all of that in mind. All except the two possibilities for the Sylow 3-subgroup. This idea of decomposing the subgroups I'm sure generalizes very widely, so this is great. Thanks.
– Alex Petzke
Feb 12 '12 at 13:15
add a comment |
By Sylow's theorem, the number of subgroups of order $5$ must divide $45$, and must be congruent to $1$ modulo $5$; the only possibility is that there is a single group of order $5$. Likewise, the number of subgroups of order $9$ must divide $45$ and be congruent to $1$ modulo $3$; the only possibility is that there is a single subgroup of order $9$.
So the group has a single subgroup of order $5$, which must be isomorphic to $mathbb{Z}_5$ (since that is the only group of order $5$), and a single subgroup of order $9$, which is isomorphic to either $mathbb{Z}_9$ or to $mathbb{Z}_3timesmathbb{Z}_3$ (since the only groups of order $p^2$, with $p$ prime, are the cyclic group of order $p^2$ and a direct product of two copies of the cyclic group of order $p$).
Moreover, both Sylow subgroups are normal; let $N$ be the $5$-Sylow subgroups, and $M$ be the $3$-Sylow subgroup. Then $Ncap M={1}$ since the orders are coprime, and $NM$ is a subgroup (since both $N$ and $M$ are normal). And
$$|NM| = frac{|N|,|M|}{|Ncap M|} = 9times 5 = 45 = |G|,$$
so $G=NM$. Since $N$ and $M$ are normal, $nm=mn$ for every $min M$ and $nin N$. So $Gcong Ntimes Mcong mathbb{Z}_5times M$. There are two possibilities for $M$, giving you the only two possibilities for $G$.
Note that in both cases you get subgroups of order $15$; what you don't get in one of the cases is an element of order $45$.
answered Feb 12 '12 at 5:27
Arturo Magidin
261k32584904
You wrote $3 times 5 = 45$. Lol
– Patrick Da Silva
Feb 12 '12 at 5:31
@Patrick: And you wrote $mathbb{Z}_3timesmathbb{Z}_3mathbb{Z}_5$... Fixed.
– Arturo Magidin
Feb 12 '12 at 5:33
Great answer. You did not assume OP's finding which was that there was only $2$ groups of order $45$, which I think is cool for people who have no idea how OP might have found that.
– Patrick Da Silva
Feb 12 '12 at 5:34
I had nearly all of that in mind. All except the two possibilities for the Sylow 3-subgroup. This idea of decomposing the subgroups I'm sure generalizes very widely, so this is great. Thanks.
– Alex Petzke
Feb 12 '12 at 13:15
add a comment |
By Sylow's theorem, the number of subgroups of order $5$ must divide $45$, and must be congruent to $1$ modulo $5$; the only possibility is that there is a single group of order $5$. Likewise, the number of subgroups of order $9$ must divide $45$ and be congruent to $1$ modulo $3$; the only possibility is that there is a single subgroup of order $9$.
So the group has a single subgroup of order $5$, which must be isomorphic to $mathbb{Z}_5$ (since that is the only group of order $5$), and a single subgroup of order $9$, which is isomorphic to either $mathbb{Z}_9$ or to $mathbb{Z}_3timesmathbb{Z}_3$ (since the only groups of order $p^2$, with $p$ prime, are the cyclic group of order $p^2$ and a direct product of two copies of the cyclic group of order $p$).
Moreover, both Sylow subgroups are normal; let $N$ be the $5$-Sylow subgroups, and $M$ be the $3$-Sylow subgroup. Then $Ncap M={1}$ since the orders are coprime, and $NM$ is a subgroup (since both $N$ and $M$ are normal). And
$$|NM| = frac{|N|,|M|}{|Ncap M|} = 9times 5 = 45 = |G|,$$
so $G=NM$. Since $N$ and $M$ are normal, $nm=mn$ for every $min M$ and $nin N$. So $Gcong Ntimes Mcong mathbb{Z}_5times M$. There are two possibilities for $M$, giving you the only two possibilities for $G$.
Note that in both cases you get subgroups of order $15$; what you don't get in one of the cases is an element of order $45$.
answered Feb 12 '12 at 5:27
Arturo Magidin
261k32584904
By Sylow's theorem, the number of subgroups of order $5$ must divide $45$, and must be congruent to $1$ modulo $5$; the only possibility is that there is a single group of order $5$. Likewise, the number of subgroups of order $9$ must divide $45$ and be congruent to $1$ modulo $3$; the only possibility is that there is a single subgroup of order $9$.
So the group has a single subgroup of order $5$, which must be isomorphic to $mathbb{Z}_5$ (since that is the only group of order $5$), and a single subgroup of order $9$, which is isomorphic to either $mathbb{Z}_9$ or to $mathbb{Z}_3timesmathbb{Z}_3$ (since the only groups of order $p^2$, with $p$ prime, are the cyclic group of order $p^2$ and a direct product of two copies of the cyclic group of order $p$).
Moreover, both Sylow subgroups are normal; let $N$ be the $5$-Sylow subgroups, and $M$ be the $3$-Sylow subgroup. Then $Ncap M={1}$ since the orders are coprime, and $NM$ is a subgroup (since both $N$ and $M$ are normal). And
$$|NM| = frac{|N|,|M|}{|Ncap M|} = 9times 5 = 45 = |G|,$$
so $G=NM$. Since $N$ and $M$ are normal, $nm=mn$ for every $min M$ and $nin N$. So $Gcong Ntimes Mcong mathbb{Z}_5times M$. There are two possibilities for $M$, giving you the only two possibilities for $G$.
Note that in both cases you get subgroups of order $15$; what you don't get in one of the cases is an element of order $45$.
answered Feb 12 '12 at 5:27
Arturo Magidin
261k32584904
answered Feb 12 '12 at 5:27
Arturo Magidin
261k32584904
answered Feb 12 '12 at 5:27
Arturo Magidin
261k32584904
answered Feb 12 '12 at 5:27
Arturo Magidin
261k32584904
261k32584904
You wrote $3 times 5 = 45$. Lol
– Patrick Da Silva
Feb 12 '12 at 5:31
@Patrick: And you wrote $mathbb{Z}_3timesmathbb{Z}_3mathbb{Z}_5$... Fixed.
– Arturo Magidin
Feb 12 '12 at 5:33
Great answer. You did not assume OP's finding which was that there was only $2$ groups of order $45$, which I think is cool for people who have no idea how OP might have found that.
– Patrick Da Silva
Feb 12 '12 at 5:34
I had nearly all of that in mind. All except the two possibilities for the Sylow 3-subgroup. This idea of decomposing the subgroups I'm sure generalizes very widely, so this is great. Thanks.
– Alex Petzke
Feb 12 '12 at 13:15
add a comment |
You wrote $3 times 5 = 45$. Lol
– Patrick Da Silva
Feb 12 '12 at 5:31
@Patrick: And you wrote $mathbb{Z}_3timesmathbb{Z}_3mathbb{Z}_5$... Fixed.
– Arturo Magidin
Feb 12 '12 at 5:33
Great answer. You did not assume OP's finding which was that there was only $2$ groups of order $45$, which I think is cool for people who have no idea how OP might have found that.
– Patrick Da Silva
Feb 12 '12 at 5:34
I had nearly all of that in mind. All except the two possibilities for the Sylow 3-subgroup. This idea of decomposing the subgroups I'm sure generalizes very widely, so this is great. Thanks.
– Alex Petzke
Feb 12 '12 at 13:15
You wrote $3 times 5 = 45$. Lol
– Patrick Da Silva
Feb 12 '12 at 5:31
You wrote $3 times 5 = 45$. Lol
– Patrick Da Silva
Feb 12 '12 at 5:31
@Patrick: And you wrote $mathbb{Z}_3timesmathbb{Z}_3mathbb{Z}_5$... Fixed.
– Arturo Magidin
Feb 12 '12 at 5:33
@Patrick: And you wrote $mathbb{Z}_3timesmathbb{Z}_3mathbb{Z}_5$... Fixed.
– Arturo Magidin
Feb 12 '12 at 5:33
Great answer. You did not assume OP's finding which was that there was only $2$ groups of order $45$, which I think is cool for people who have no idea how OP might have found that.
– Patrick Da Silva
Feb 12 '12 at 5:34
Great answer. You did not assume OP's finding which was that there was only $2$ groups of order $45$, which I think is cool for people who have no idea how OP might have found that.
– Patrick Da Silva
Feb 12 '12 at 5:34
I had nearly all of that in mind. All except the two possibilities for the Sylow 3-subgroup. This idea of decomposing the subgroups I'm sure generalizes very widely, so this is great. Thanks.
– Alex Petzke
Feb 12 '12 at 13:15
I had nearly all of that in mind. All except the two possibilities for the Sylow 3-subgroup. This idea of decomposing the subgroups I'm sure generalizes very widely, so this is great. Thanks.
– Alex Petzke
Feb 12 '12 at 13:15
add a comment |
If you know that there are two groups of order $45$, the first groups you must think of are abelian groups, i.e. direct product of cyclic groups by the finitely-generated-abelian-groups classification. Now your "building blocks" are $mathbb Z_3$, which you have twice, and $mathbb Z_5$, which you have once. If you try to "put them all together" you get $mathbb Z_{45}$, which we clearly expected. Now other combinations would be $mathbb Z_9 times mathbb Z_5$, $mathbb Z_3 times mathbb Z_3times mathbb Z_5$, and $mathbb Z_{15} times mathbb Z_3$. For the first one, since $g.c.d.(9,5) = 1$, you know that this group will be cyclic, hence isomorphic to $mathbb Z_{45}$.
For the other two, since $g.c.d(3,5) = 1$, $mathbb Z_3 times mathbb Z_5$ is isomorphic to $mathbb Z_{15}$ so we don't have anything new there. But in $mathbb Z_{15} times mathbb Z_3$, every element has order that divides $15$, thus this group cannot be isomorphic to $mathbb Z_{45}$.
Hope that helps,
edited Feb 12 '12 at 5:32
Arturo Magidin
261k32584904
answered Feb 12 '12 at 5:29
Patrick Da Silva
32k353106
Heh you were quicker than me on that one!
– Patrick Da Silva
Feb 12 '12 at 5:33
Yes, knowing that it was abelian I should have followed up with the Fundamental Theorem of Finite Abelian Groups... Another good way of looking at it. Thanks.
– Alex Petzke
Feb 12 '12 at 20:22
add a comment |
If you know that there are two groups of order $45$, the first groups you must think of are abelian groups, i.e. direct product of cyclic groups by the finitely-generated-abelian-groups classification. Now your "building blocks" are $mathbb Z_3$, which you have twice, and $mathbb Z_5$, which you have once. If you try to "put them all together" you get $mathbb Z_{45}$, which we clearly expected. Now other combinations would be $mathbb Z_9 times mathbb Z_5$, $mathbb Z_3 times mathbb Z_3times mathbb Z_5$, and $mathbb Z_{15} times mathbb Z_3$. For the first one, since $g.c.d.(9,5) = 1$, you know that this group will be cyclic, hence isomorphic to $mathbb Z_{45}$.
For the other two, since $g.c.d(3,5) = 1$, $mathbb Z_3 times mathbb Z_5$ is isomorphic to $mathbb Z_{15}$ so we don't have anything new there. But in $mathbb Z_{15} times mathbb Z_3$, every element has order that divides $15$, thus this group cannot be isomorphic to $mathbb Z_{45}$.
Hope that helps,
edited Feb 12 '12 at 5:32
Arturo Magidin
261k32584904
answered Feb 12 '12 at 5:29
Patrick Da Silva
32k353106
Heh you were quicker than me on that one!
– Patrick Da Silva
Feb 12 '12 at 5:33
Yes, knowing that it was abelian I should have followed up with the Fundamental Theorem of Finite Abelian Groups... Another good way of looking at it. Thanks.
– Alex Petzke
Feb 12 '12 at 20:22
add a comment |
If you know that there are two groups of order $45$, the first groups you must think of are abelian groups, i.e. direct product of cyclic groups by the finitely-generated-abelian-groups classification. Now your "building blocks" are $mathbb Z_3$, which you have twice, and $mathbb Z_5$, which you have once. If you try to "put them all together" you get $mathbb Z_{45}$, which we clearly expected. Now other combinations would be $mathbb Z_9 times mathbb Z_5$, $mathbb Z_3 times mathbb Z_3times mathbb Z_5$, and $mathbb Z_{15} times mathbb Z_3$. For the first one, since $g.c.d.(9,5) = 1$, you know that this group will be cyclic, hence isomorphic to $mathbb Z_{45}$.
For the other two, since $g.c.d(3,5) = 1$, $mathbb Z_3 times mathbb Z_5$ is isomorphic to $mathbb Z_{15}$ so we don't have anything new there. But in $mathbb Z_{15} times mathbb Z_3$, every element has order that divides $15$, thus this group cannot be isomorphic to $mathbb Z_{45}$.
Hope that helps,
edited Feb 12 '12 at 5:32
Arturo Magidin
261k32584904
answered Feb 12 '12 at 5:29
Patrick Da Silva
32k353106
If you know that there are two groups of order $45$, the first groups you must think of are abelian groups, i.e. direct product of cyclic groups by the finitely-generated-abelian-groups classification. Now your "building blocks" are $mathbb Z_3$, which you have twice, and $mathbb Z_5$, which you have once. If you try to "put them all together" you get $mathbb Z_{45}$, which we clearly expected. Now other combinations would be $mathbb Z_9 times mathbb Z_5$, $mathbb Z_3 times mathbb Z_3times mathbb Z_5$, and $mathbb Z_{15} times mathbb Z_3$. For the first one, since $g.c.d.(9,5) = 1$, you know that this group will be cyclic, hence isomorphic to $mathbb Z_{45}$.
For the other two, since $g.c.d(3,5) = 1$, $mathbb Z_3 times mathbb Z_5$ is isomorphic to $mathbb Z_{15}$ so we don't have anything new there. But in $mathbb Z_{15} times mathbb Z_3$, every element has order that divides $15$, thus this group cannot be isomorphic to $mathbb Z_{45}$.
Hope that helps,
edited Feb 12 '12 at 5:32
Arturo Magidin
261k32584904
answered Feb 12 '12 at 5:29
Patrick Da Silva
32k353106
edited Feb 12 '12 at 5:32
Arturo Magidin
261k32584904
edited Feb 12 '12 at 5:32
Arturo Magidin
261k32584904
edited Feb 12 '12 at 5:32
Arturo Magidin
261k32584904
261k32584904
answered Feb 12 '12 at 5:29
Patrick Da Silva
32k353106
answered Feb 12 '12 at 5:29
Patrick Da Silva
32k353106
answered Feb 12 '12 at 5:29
Patrick Da Silva
32k353106
32k353106
Heh you were quicker than me on that one!
– Patrick Da Silva
Feb 12 '12 at 5:33
Yes, knowing that it was abelian I should have followed up with the Fundamental Theorem of Finite Abelian Groups... Another good way of looking at it. Thanks.
– Alex Petzke
Feb 12 '12 at 20:22
add a comment |
Heh you were quicker than me on that one!
– Patrick Da Silva
Feb 12 '12 at 5:33
Yes, knowing that it was abelian I should have followed up with the Fundamental Theorem of Finite Abelian Groups... Another good way of looking at it. Thanks.
– Alex Petzke
Feb 12 '12 at 20:22
Heh you were quicker than me on that one!
– Patrick Da Silva
Feb 12 '12 at 5:33
Heh you were quicker than me on that one!
– Patrick Da Silva
Feb 12 '12 at 5:33
Yes, knowing that it was abelian I should have followed up with the Fundamental Theorem of Finite Abelian Groups... Another good way of looking at it. Thanks.
– Alex Petzke
Feb 12 '12 at 20:22
Yes, knowing that it was abelian I should have followed up with the Fundamental Theorem of Finite Abelian Groups... Another good way of looking at it. Thanks.
– Alex Petzke
Feb 12 '12 at 20:22
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It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.
answered Feb 12 '12 at 5:23
Jim Belk
37.4k285150
add a comment |
It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.
answered Feb 12 '12 at 5:23
Jim Belk
37.4k285150
add a comment |
It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.
answered Feb 12 '12 at 5:23
Jim Belk
37.4k285150
It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.
answered Feb 12 '12 at 5:23
Jim Belk
37.4k285150
answered Feb 12 '12 at 5:23
Jim Belk
37.4k285150
answered Feb 12 '12 at 5:23
Jim Belk
37.4k285150
answered Feb 12 '12 at 5:23
Jim Belk
37.4k285150
37.4k285150
add a comment |
add a comment |
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