Every nonabelian group of order 6 has a non-normal subgroup of order 2 (revisited)












7














I am fully aware that this question has already been addressed here and here. The question, however, derives from a Dummit and Foote exercise (section 4.2 exercise 10 page 122) and the answers provided make use of material that appears later in the book, or, at least, I do not clearly understand them in terms of the material that I have studied already.



So, I would like to submit the following tentative proof
'from first principles' (i.e. Dummit and Foote before page 122). As I feel insecure about it, I would be grateful if you could check it.



Consider a nonabelian group G of order 6. By Cauchy's theorem, the group contains at least one element of order 2, and therefore at least one subgroup of order 2. Suppose this/these subgroups are all normal. This would imply that all elements of order 2 commute with all elements of G.



The remaining elements of group G are of order 1 (which trivially is in the center of G), or order 3 (order 6 would imply the group is cyclic and therefore abelian). Since all elements of order 2 are in the center, the order 3 elements will also be in the center of G.



Therefore, G would be abelian, contrary to the assumption. So, at least one of the subgroups of order 2 should be non-normal.






abstract-algebra group-theory







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  • Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
    – Berci
    43 mins ago
















7














I am fully aware that this question has already been addressed here and here. The question, however, derives from a Dummit and Foote exercise (section 4.2 exercise 10 page 122) and the answers provided make use of material that appears later in the book, or, at least, I do not clearly understand them in terms of the material that I have studied already.



So, I would like to submit the following tentative proof
'from first principles' (i.e. Dummit and Foote before page 122). As I feel insecure about it, I would be grateful if you could check it.



Consider a nonabelian group G of order 6. By Cauchy's theorem, the group contains at least one element of order 2, and therefore at least one subgroup of order 2. Suppose this/these subgroups are all normal. This would imply that all elements of order 2 commute with all elements of G.



The remaining elements of group G are of order 1 (which trivially is in the center of G), or order 3 (order 6 would imply the group is cyclic and therefore abelian). Since all elements of order 2 are in the center, the order 3 elements will also be in the center of G.



Therefore, G would be abelian, contrary to the assumption. So, at least one of the subgroups of order 2 should be non-normal.






abstract-algebra group-theory







share|cite|improve this question






















  • Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
    – Berci
    43 mins ago














7












7








7







I am fully aware that this question has already been addressed here and here. The question, however, derives from a Dummit and Foote exercise (section 4.2 exercise 10 page 122) and the answers provided make use of material that appears later in the book, or, at least, I do not clearly understand them in terms of the material that I have studied already.



So, I would like to submit the following tentative proof
'from first principles' (i.e. Dummit and Foote before page 122). As I feel insecure about it, I would be grateful if you could check it.



Consider a nonabelian group G of order 6. By Cauchy's theorem, the group contains at least one element of order 2, and therefore at least one subgroup of order 2. Suppose this/these subgroups are all normal. This would imply that all elements of order 2 commute with all elements of G.



The remaining elements of group G are of order 1 (which trivially is in the center of G), or order 3 (order 6 would imply the group is cyclic and therefore abelian). Since all elements of order 2 are in the center, the order 3 elements will also be in the center of G.



Therefore, G would be abelian, contrary to the assumption. So, at least one of the subgroups of order 2 should be non-normal.






abstract-algebra group-theory







share|cite|improve this question













I am fully aware that this question has already been addressed here and here. The question, however, derives from a Dummit and Foote exercise (section 4.2 exercise 10 page 122) and the answers provided make use of material that appears later in the book, or, at least, I do not clearly understand them in terms of the material that I have studied already.



So, I would like to submit the following tentative proof
'from first principles' (i.e. Dummit and Foote before page 122). As I feel insecure about it, I would be grateful if you could check it.



Consider a nonabelian group G of order 6. By Cauchy's theorem, the group contains at least one element of order 2, and therefore at least one subgroup of order 2. Suppose this/these subgroups are all normal. This would imply that all elements of order 2 commute with all elements of G.



The remaining elements of group G are of order 1 (which trivially is in the center of G), or order 3 (order 6 would imply the group is cyclic and therefore abelian). Since all elements of order 2 are in the center, the order 3 elements will also be in the center of G.



Therefore, G would be abelian, contrary to the assumption. So, at least one of the subgroups of order 2 should be non-normal.






abstract-algebra group-theory




abstract-algebra group-theory






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asked 2 hours ago









Frank De Geeter

604




604












  • Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
    – Berci
    43 mins ago


















  • Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
    – Berci
    43 mins ago
















Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
– Berci
43 mins ago




Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
– Berci
43 mins ago










2 Answers
2






active

oldest

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3














Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.






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    1














    Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
    $$
    r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
    $$
    so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.






    share|cite|improve this answer





















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      2 Answers
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      active

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      3














      Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.






      share|cite|improve this answer


























        3














        Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.






        share|cite|improve this answer
























          3












          3








          3






          Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.






          share|cite|improve this answer












          Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Nicky Hekster

          28.2k53456




          28.2k53456























              1














              Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
              $$
              r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
              $$
              so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.






              share|cite|improve this answer


























                1














                Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
                $$
                r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
                $$
                so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
                  $$
                  r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
                  $$
                  so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.






                  share|cite|improve this answer












                  Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
                  $$
                  r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
                  $$
                  so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 50 mins ago









                  Dietrich Burde

                  77.8k64386




                  77.8k64386






























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