Finite group has a generating set

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Let $G$ be a finite group. Show that there exists a generating set $S$ of $G$ with $|S| le log_2 |G|$.




I started looking what can I get if I look at generating sets of elements of $G$,i.e., $<x_1>,<x_1,x_2>,<x_1,x_2,x_3>...$ and yet I'm not sure how can I limit it as required.



Any other ideas?




group-theory finite-groups




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    up vote
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    down vote

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    Let $G$ be a finite group. Show that there exists a generating set $S$ of $G$ with $|S| le log_2 |G|$.




    I started looking what can I get if I look at generating sets of elements of $G$,i.e., $<x_1>,<x_1,x_2>,<x_1,x_2,x_3>...$ and yet I'm not sure how can I limit it as required.



    Any other ideas?




    group-theory finite-groups




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Let $G$ be a finite group. Show that there exists a generating set $S$ of $G$ with $|S| le log_2 |G|$.




      I started looking what can I get if I look at generating sets of elements of $G$,i.e., $<x_1>,<x_1,x_2>,<x_1,x_2,x_3>...$ and yet I'm not sure how can I limit it as required.



      Any other ideas?




      group-theory finite-groups




      share|cite|improve this question












      Let $G$ be a finite group. Show that there exists a generating set $S$ of $G$ with $|S| le log_2 |G|$.




      I started looking what can I get if I look at generating sets of elements of $G$,i.e., $<x_1>,<x_1,x_2>,<x_1,x_2,x_3>...$ and yet I'm not sure how can I limit it as required.



      Any other ideas?





      group-theory finite-groups





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 24 at 20:09









      ChikChak

      659214




      659214




















          1 Answer
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          Basic idea: simply pick elements successively so that the next one is not in the generated subgroup of the elements you already picked. In each step, the size of the generated subgroup at least doubles.
          This solves the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$. To improve this, we break the problem to two cases.



          Case 1: If every element has order 2 or 3 in $G$. Then by Cauchy's theorem, $|G|= 2^k3^m$. By Burnside's theorem, $G$ is solvable:



          https://en.wikipedia.org/wiki/Burnside_theorem



          So there is a sequence of groups, each an extension of the previous one by $mathbbZ_2$ or $mathbbZ_3$. Thus we can pick the new elements successively so that in the i-th step we generate the i-th subgroup in the chain. This requires $k+m$ elements, which is clearly strictly less than $log_2 |G|$.



          Case 2: There is an element $g$ whose order is at least 4. Then start by this. So now instead of having the lower estimation for the size of the i-th subgroup $2^i$ (see the basic idea), we have $2^i+1$, as the first generated subgroup has size at least $4=2^2$. This is just enough, the number of elements is at most the upper floor of $log_2 |G|$ minus 1 now.






          share|cite|improve this answer























          • Which is what I tried, but I can't say what's the size of the first generating set to begin with.
            – ChikChak
            Jul 24 at 20:15






          • 1




            If G is trivial, you can generate with zero elements, which is good. If it is non-trivial, then pick any element different from the identity as a first one. If fact, the first step is no different from the others. The 1-element subgroup is generated by default, so the same rule applies: pick one that is not in the subgroup generated by the elements you already picked (no elements picked, so the generated subgroup is the trivial one).
            – A. Pongrácz
            Jul 24 at 20:19







          • 1




            I understood you, and I answered your question. It is irrelevant. The only important thing is that is is at least two (if you pick a nontrivial element first).
            – A. Pongrácz
            Jul 24 at 20:23






          • 1




            I already have.
            – A. Pongrácz
            Jul 24 at 20:38






          • 1




            Is your point that I only managed to solve the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$, instead of lower floor? I can fix that.
            – A. Pongrácz
            Jul 24 at 20:46











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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Basic idea: simply pick elements successively so that the next one is not in the generated subgroup of the elements you already picked. In each step, the size of the generated subgroup at least doubles.
          This solves the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$. To improve this, we break the problem to two cases.



          Case 1: If every element has order 2 or 3 in $G$. Then by Cauchy's theorem, $|G|= 2^k3^m$. By Burnside's theorem, $G$ is solvable:



          https://en.wikipedia.org/wiki/Burnside_theorem



          So there is a sequence of groups, each an extension of the previous one by $mathbbZ_2$ or $mathbbZ_3$. Thus we can pick the new elements successively so that in the i-th step we generate the i-th subgroup in the chain. This requires $k+m$ elements, which is clearly strictly less than $log_2 |G|$.



          Case 2: There is an element $g$ whose order is at least 4. Then start by this. So now instead of having the lower estimation for the size of the i-th subgroup $2^i$ (see the basic idea), we have $2^i+1$, as the first generated subgroup has size at least $4=2^2$. This is just enough, the number of elements is at most the upper floor of $log_2 |G|$ minus 1 now.






          share|cite|improve this answer























          • Which is what I tried, but I can't say what's the size of the first generating set to begin with.
            – ChikChak
            Jul 24 at 20:15






          • 1




            If G is trivial, you can generate with zero elements, which is good. If it is non-trivial, then pick any element different from the identity as a first one. If fact, the first step is no different from the others. The 1-element subgroup is generated by default, so the same rule applies: pick one that is not in the subgroup generated by the elements you already picked (no elements picked, so the generated subgroup is the trivial one).
            – A. Pongrácz
            Jul 24 at 20:19







          • 1




            I understood you, and I answered your question. It is irrelevant. The only important thing is that is is at least two (if you pick a nontrivial element first).
            – A. Pongrácz
            Jul 24 at 20:23






          • 1




            I already have.
            – A. Pongrácz
            Jul 24 at 20:38






          • 1




            Is your point that I only managed to solve the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$, instead of lower floor? I can fix that.
            – A. Pongrácz
            Jul 24 at 20:46















          up vote
          1
          down vote



          accepted










          Basic idea: simply pick elements successively so that the next one is not in the generated subgroup of the elements you already picked. In each step, the size of the generated subgroup at least doubles.
          This solves the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$. To improve this, we break the problem to two cases.



          Case 1: If every element has order 2 or 3 in $G$. Then by Cauchy's theorem, $|G|= 2^k3^m$. By Burnside's theorem, $G$ is solvable:



          https://en.wikipedia.org/wiki/Burnside_theorem



          So there is a sequence of groups, each an extension of the previous one by $mathbbZ_2$ or $mathbbZ_3$. Thus we can pick the new elements successively so that in the i-th step we generate the i-th subgroup in the chain. This requires $k+m$ elements, which is clearly strictly less than $log_2 |G|$.



          Case 2: There is an element $g$ whose order is at least 4. Then start by this. So now instead of having the lower estimation for the size of the i-th subgroup $2^i$ (see the basic idea), we have $2^i+1$, as the first generated subgroup has size at least $4=2^2$. This is just enough, the number of elements is at most the upper floor of $log_2 |G|$ minus 1 now.






          share|cite|improve this answer























          • Which is what I tried, but I can't say what's the size of the first generating set to begin with.
            – ChikChak
            Jul 24 at 20:15






          • 1




            If G is trivial, you can generate with zero elements, which is good. If it is non-trivial, then pick any element different from the identity as a first one. If fact, the first step is no different from the others. The 1-element subgroup is generated by default, so the same rule applies: pick one that is not in the subgroup generated by the elements you already picked (no elements picked, so the generated subgroup is the trivial one).
            – A. Pongrácz
            Jul 24 at 20:19







          • 1




            I understood you, and I answered your question. It is irrelevant. The only important thing is that is is at least two (if you pick a nontrivial element first).
            – A. Pongrácz
            Jul 24 at 20:23






          • 1




            I already have.
            – A. Pongrácz
            Jul 24 at 20:38






          • 1




            Is your point that I only managed to solve the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$, instead of lower floor? I can fix that.
            – A. Pongrácz
            Jul 24 at 20:46













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Basic idea: simply pick elements successively so that the next one is not in the generated subgroup of the elements you already picked. In each step, the size of the generated subgroup at least doubles.
          This solves the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$. To improve this, we break the problem to two cases.



          Case 1: If every element has order 2 or 3 in $G$. Then by Cauchy's theorem, $|G|= 2^k3^m$. By Burnside's theorem, $G$ is solvable:



          https://en.wikipedia.org/wiki/Burnside_theorem



          So there is a sequence of groups, each an extension of the previous one by $mathbbZ_2$ or $mathbbZ_3$. Thus we can pick the new elements successively so that in the i-th step we generate the i-th subgroup in the chain. This requires $k+m$ elements, which is clearly strictly less than $log_2 |G|$.



          Case 2: There is an element $g$ whose order is at least 4. Then start by this. So now instead of having the lower estimation for the size of the i-th subgroup $2^i$ (see the basic idea), we have $2^i+1$, as the first generated subgroup has size at least $4=2^2$. This is just enough, the number of elements is at most the upper floor of $log_2 |G|$ minus 1 now.






          share|cite|improve this answer















          Basic idea: simply pick elements successively so that the next one is not in the generated subgroup of the elements you already picked. In each step, the size of the generated subgroup at least doubles.
          This solves the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$. To improve this, we break the problem to two cases.



          Case 1: If every element has order 2 or 3 in $G$. Then by Cauchy's theorem, $|G|= 2^k3^m$. By Burnside's theorem, $G$ is solvable:



          https://en.wikipedia.org/wiki/Burnside_theorem



          So there is a sequence of groups, each an extension of the previous one by $mathbbZ_2$ or $mathbbZ_3$. Thus we can pick the new elements successively so that in the i-th step we generate the i-th subgroup in the chain. This requires $k+m$ elements, which is clearly strictly less than $log_2 |G|$.



          Case 2: There is an element $g$ whose order is at least 4. Then start by this. So now instead of having the lower estimation for the size of the i-th subgroup $2^i$ (see the basic idea), we have $2^i+1$, as the first generated subgroup has size at least $4=2^2$. This is just enough, the number of elements is at most the upper floor of $log_2 |G|$ minus 1 now.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 24 at 20:54


























          answered Jul 24 at 20:13









          A. Pongrácz

          1,804116




          1,804116











          • Which is what I tried, but I can't say what's the size of the first generating set to begin with.
            – ChikChak
            Jul 24 at 20:15






          • 1




            If G is trivial, you can generate with zero elements, which is good. If it is non-trivial, then pick any element different from the identity as a first one. If fact, the first step is no different from the others. The 1-element subgroup is generated by default, so the same rule applies: pick one that is not in the subgroup generated by the elements you already picked (no elements picked, so the generated subgroup is the trivial one).
            – A. Pongrácz
            Jul 24 at 20:19







          • 1




            I understood you, and I answered your question. It is irrelevant. The only important thing is that is is at least two (if you pick a nontrivial element first).
            – A. Pongrácz
            Jul 24 at 20:23






          • 1




            I already have.
            – A. Pongrácz
            Jul 24 at 20:38






          • 1




            Is your point that I only managed to solve the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$, instead of lower floor? I can fix that.
            – A. Pongrácz
            Jul 24 at 20:46

















          • Which is what I tried, but I can't say what's the size of the first generating set to begin with.
            – ChikChak
            Jul 24 at 20:15






          • 1




            If G is trivial, you can generate with zero elements, which is good. If it is non-trivial, then pick any element different from the identity as a first one. If fact, the first step is no different from the others. The 1-element subgroup is generated by default, so the same rule applies: pick one that is not in the subgroup generated by the elements you already picked (no elements picked, so the generated subgroup is the trivial one).
            – A. Pongrácz
            Jul 24 at 20:19







          • 1




            I understood you, and I answered your question. It is irrelevant. The only important thing is that is is at least two (if you pick a nontrivial element first).
            – A. Pongrácz
            Jul 24 at 20:23






          • 1




            I already have.
            – A. Pongrácz
            Jul 24 at 20:38






          • 1




            Is your point that I only managed to solve the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$, instead of lower floor? I can fix that.
            – A. Pongrácz
            Jul 24 at 20:46
















          Which is what I tried, but I can't say what's the size of the first generating set to begin with.
          – ChikChak
          Jul 24 at 20:15




          Which is what I tried, but I can't say what's the size of the first generating set to begin with.
          – ChikChak
          Jul 24 at 20:15




          1




          1




          If G is trivial, you can generate with zero elements, which is good. If it is non-trivial, then pick any element different from the identity as a first one. If fact, the first step is no different from the others. The 1-element subgroup is generated by default, so the same rule applies: pick one that is not in the subgroup generated by the elements you already picked (no elements picked, so the generated subgroup is the trivial one).
          – A. Pongrácz
          Jul 24 at 20:19





          If G is trivial, you can generate with zero elements, which is good. If it is non-trivial, then pick any element different from the identity as a first one. If fact, the first step is no different from the others. The 1-element subgroup is generated by default, so the same rule applies: pick one that is not in the subgroup generated by the elements you already picked (no elements picked, so the generated subgroup is the trivial one).
          – A. Pongrácz
          Jul 24 at 20:19





          1




          1




          I understood you, and I answered your question. It is irrelevant. The only important thing is that is is at least two (if you pick a nontrivial element first).
          – A. Pongrácz
          Jul 24 at 20:23




          I understood you, and I answered your question. It is irrelevant. The only important thing is that is is at least two (if you pick a nontrivial element first).
          – A. Pongrácz
          Jul 24 at 20:23




          1




          1




          I already have.
          – A. Pongrácz
          Jul 24 at 20:38




          I already have.
          – A. Pongrácz
          Jul 24 at 20:38




          1




          1




          Is your point that I only managed to solve the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$, instead of lower floor? I can fix that.
          – A. Pongrácz
          Jul 24 at 20:46





          Is your point that I only managed to solve the problem with $n$ elements where $n$ is the upper floor of $log_2 |G|$, instead of lower floor? I can fix that.
          – A. Pongrácz
          Jul 24 at 20:46













           

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