Explicit descriptions of groups of order 45











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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?










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    up vote
    13
    down vote

    favorite
    3












    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?










    share|cite|improve this question


























      up vote
      13
      down vote

      favorite
      3









      up vote
      13
      down vote

      favorite
      3






      3





      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?










      share|cite|improve this question















      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






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 12 '12 at 5:21









      Arturo Magidin

      260k32581902




      260k32581902










      asked Feb 12 '12 at 5:19









      Alex Petzke

      3,95923569




      3,95923569






















          3 Answers
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          up vote
          17
          down vote



          accepted










          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 $45$.






          share|cite|improve this answer





















          • 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


















          up vote
          6
          down vote













          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,






          share|cite|improve this answer























          • 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


















          up vote
          1
          down vote













          It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.






          share|cite|improve this answer





















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            3 Answers
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            3 Answers
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            active

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            up vote
            17
            down vote



            accepted










            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 $45$.






            share|cite|improve this answer





















            • 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















            up vote
            17
            down vote



            accepted










            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 $45$.






            share|cite|improve this answer





















            • 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













            up vote
            17
            down vote



            accepted







            up vote
            17
            down vote



            accepted






            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 $45$.






            share|cite|improve this answer












            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 $45$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 12 '12 at 5:27









            Arturo Magidin

            260k32581902




            260k32581902












            • 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










            • @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










            up vote
            6
            down vote













            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,






            share|cite|improve this answer























            • 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















            up vote
            6
            down vote













            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,






            share|cite|improve this answer























            • 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













            up vote
            6
            down vote










            up vote
            6
            down vote









            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,






            share|cite|improve this answer














            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,







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 12 '12 at 5:32









            Arturo Magidin

            260k32581902




            260k32581902










            answered Feb 12 '12 at 5:29









            Patrick Da Silva

            31.9k353106




            31.9k353106












            • 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










            • 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










            up vote
            1
            down vote













            It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.






            share|cite|improve this answer

























              up vote
              1
              down vote













              It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.






                share|cite|improve this answer












                It's $mathbb{Z}_{15} times mathbb{Z}_3$. Note that there is indeed a subgroup of order $15$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 12 '12 at 5:23









                Jim Belk

                37.3k284150




                37.3k284150






























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