Each group of order 8 has a subgroup of order 2 and a subgroup of order 4.












6












$begingroup$


So, I was trying to prove the following theorem:




Let $G$ be a group of order $8$. So $G$ has a subgroup of order $2$ and a subgroup of order $4$.




First I proved that if a group has a finite even order, it has an element $g_0$ of order $2$. So $H={g_0,e}$ is a group of order $2$. Now I'm trying to find the group of order $4$. By Lagrange theorem I can understand that:



$$|G|=|H|cdot |G,:,H| Rightarrow |G,:,H|=4$$



So we have $4$ Cosets. But how to continue from here? I feel like I miss the last one-two lines of the proof.






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you know that nilpotent groups (and in particular $p$-groups) have non-trivial centres? If yes, it would be better and more instructive to prove that nilpotent groups (and in particular $p$-groups) have subgroups of each possible order. (In fact, this holds for a broader class of groups.)
    $endgroup$
    – the_fox
    Feb 15 at 17:37
















6












$begingroup$


So, I was trying to prove the following theorem:




Let $G$ be a group of order $8$. So $G$ has a subgroup of order $2$ and a subgroup of order $4$.




First I proved that if a group has a finite even order, it has an element $g_0$ of order $2$. So $H={g_0,e}$ is a group of order $2$. Now I'm trying to find the group of order $4$. By Lagrange theorem I can understand that:



$$|G|=|H|cdot |G,:,H| Rightarrow |G,:,H|=4$$



So we have $4$ Cosets. But how to continue from here? I feel like I miss the last one-two lines of the proof.






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you know that nilpotent groups (and in particular $p$-groups) have non-trivial centres? If yes, it would be better and more instructive to prove that nilpotent groups (and in particular $p$-groups) have subgroups of each possible order. (In fact, this holds for a broader class of groups.)
    $endgroup$
    – the_fox
    Feb 15 at 17:37














6












6








6





$begingroup$


So, I was trying to prove the following theorem:




Let $G$ be a group of order $8$. So $G$ has a subgroup of order $2$ and a subgroup of order $4$.




First I proved that if a group has a finite even order, it has an element $g_0$ of order $2$. So $H={g_0,e}$ is a group of order $2$. Now I'm trying to find the group of order $4$. By Lagrange theorem I can understand that:



$$|G|=|H|cdot |G,:,H| Rightarrow |G,:,H|=4$$



So we have $4$ Cosets. But how to continue from here? I feel like I miss the last one-two lines of the proof.






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




So, I was trying to prove the following theorem:




Let $G$ be a group of order $8$. So $G$ has a subgroup of order $2$ and a subgroup of order $4$.




First I proved that if a group has a finite even order, it has an element $g_0$ of order $2$. So $H={g_0,e}$ is a group of order $2$. Now I'm trying to find the group of order $4$. By Lagrange theorem I can understand that:



$$|G|=|H|cdot |G,:,H| Rightarrow |G,:,H|=4$$



So we have $4$ Cosets. But how to continue from here? I feel like I miss the last one-two lines of the proof.






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 15 at 17:58









Shaun

9,268113684




9,268113684










asked Feb 15 at 17:23









BadukBaduk

493




493








  • 1




    $begingroup$
    Do you know that nilpotent groups (and in particular $p$-groups) have non-trivial centres? If yes, it would be better and more instructive to prove that nilpotent groups (and in particular $p$-groups) have subgroups of each possible order. (In fact, this holds for a broader class of groups.)
    $endgroup$
    – the_fox
    Feb 15 at 17:37














  • 1




    $begingroup$
    Do you know that nilpotent groups (and in particular $p$-groups) have non-trivial centres? If yes, it would be better and more instructive to prove that nilpotent groups (and in particular $p$-groups) have subgroups of each possible order. (In fact, this holds for a broader class of groups.)
    $endgroup$
    – the_fox
    Feb 15 at 17:37








1




1




$begingroup$
Do you know that nilpotent groups (and in particular $p$-groups) have non-trivial centres? If yes, it would be better and more instructive to prove that nilpotent groups (and in particular $p$-groups) have subgroups of each possible order. (In fact, this holds for a broader class of groups.)
$endgroup$
– the_fox
Feb 15 at 17:37




$begingroup$
Do you know that nilpotent groups (and in particular $p$-groups) have non-trivial centres? If yes, it would be better and more instructive to prove that nilpotent groups (and in particular $p$-groups) have subgroups of each possible order. (In fact, this holds for a broader class of groups.)
$endgroup$
– the_fox
Feb 15 at 17:37










3 Answers
3






active

oldest

votes


















8












$begingroup$

Well, you can show that $G$ has a subgroup of order 4 directly. If $G$ has an element of order 4 or 8 then we are already done [why?]



Otherwise let $pi$ and $alpha$ be two distinct elements in $G$ of order 2. So $pi = pi^{-1}$ and $alpha = alpha^{-1}$. Then $pialpha$ must have order 2 as well, lest there be an element of order 4 or 8 which would imply that we are done. This implies that $pi alpha = alpha pi$. [Indeed, for any elements $a,b,c$ in any group $G'$, if $ab=ac$ then $b$ must equal $c$. But here $(pi alpha)(pi alpha) = 1$, while $(pi alpha)(alpha pi) = (pi alpha)(alpha^{-1}pi^{-1}) = 1$. So indeed, $pi alpha = alpha pi$.] This implies that $H doteq {1,alpha, pi, pialpha }$ is closed under composition, and as every element in $H$ has its inverse in $H$, it follows that $H$ is a subgroup, and has order 4.






share|cite|improve this answer











$endgroup$





















    6












    $begingroup$

    If $G$ has no subgroup of order $4$, then every $gne e$ has order $2$. Then for any $g,h$,
    $$
    e=(gh)^2=ghgh=ghg^{-1}h^{-1},
    $$    i.e. $G$ is abelian. Then ${e,g,h,gh}$ would be a subgroup of order $4$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      why every $gneq e$ has order $2$ if $G$ does not have subgroup of order $4$? Also why $ghgh=ghg^{-1}h^{-1}$?
      $endgroup$
      – Baduk
      Feb 15 at 18:21












    • $begingroup$
      @Baduk: Apply Lagrange's Theorem. Also, we have $x^{-1}=x$ iff $x^2=e$. (Why?)
      $endgroup$
      – Shaun
      Feb 15 at 18:50





















    3












    $begingroup$

    Hint: No group has exactly two elements of order two.




    Assume no subgroup of order four exists. Apply Lagrange's Theorem. Apply the hint. See @Mike's answer.




    Reference for the lemma that is the hint:



    Exercise 4.61 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)"






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
      $endgroup$
      – Gibbs
      Feb 15 at 17:58






    • 1




      $begingroup$
      @Gibbs, I've included a reference.
      $endgroup$
      – Shaun
      Feb 15 at 18:01











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    3 Answers
    3






    active

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    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    Well, you can show that $G$ has a subgroup of order 4 directly. If $G$ has an element of order 4 or 8 then we are already done [why?]



    Otherwise let $pi$ and $alpha$ be two distinct elements in $G$ of order 2. So $pi = pi^{-1}$ and $alpha = alpha^{-1}$. Then $pialpha$ must have order 2 as well, lest there be an element of order 4 or 8 which would imply that we are done. This implies that $pi alpha = alpha pi$. [Indeed, for any elements $a,b,c$ in any group $G'$, if $ab=ac$ then $b$ must equal $c$. But here $(pi alpha)(pi alpha) = 1$, while $(pi alpha)(alpha pi) = (pi alpha)(alpha^{-1}pi^{-1}) = 1$. So indeed, $pi alpha = alpha pi$.] This implies that $H doteq {1,alpha, pi, pialpha }$ is closed under composition, and as every element in $H$ has its inverse in $H$, it follows that $H$ is a subgroup, and has order 4.






    share|cite|improve this answer











    $endgroup$


















      8












      $begingroup$

      Well, you can show that $G$ has a subgroup of order 4 directly. If $G$ has an element of order 4 or 8 then we are already done [why?]



      Otherwise let $pi$ and $alpha$ be two distinct elements in $G$ of order 2. So $pi = pi^{-1}$ and $alpha = alpha^{-1}$. Then $pialpha$ must have order 2 as well, lest there be an element of order 4 or 8 which would imply that we are done. This implies that $pi alpha = alpha pi$. [Indeed, for any elements $a,b,c$ in any group $G'$, if $ab=ac$ then $b$ must equal $c$. But here $(pi alpha)(pi alpha) = 1$, while $(pi alpha)(alpha pi) = (pi alpha)(alpha^{-1}pi^{-1}) = 1$. So indeed, $pi alpha = alpha pi$.] This implies that $H doteq {1,alpha, pi, pialpha }$ is closed under composition, and as every element in $H$ has its inverse in $H$, it follows that $H$ is a subgroup, and has order 4.






      share|cite|improve this answer











      $endgroup$
















        8












        8








        8





        $begingroup$

        Well, you can show that $G$ has a subgroup of order 4 directly. If $G$ has an element of order 4 or 8 then we are already done [why?]



        Otherwise let $pi$ and $alpha$ be two distinct elements in $G$ of order 2. So $pi = pi^{-1}$ and $alpha = alpha^{-1}$. Then $pialpha$ must have order 2 as well, lest there be an element of order 4 or 8 which would imply that we are done. This implies that $pi alpha = alpha pi$. [Indeed, for any elements $a,b,c$ in any group $G'$, if $ab=ac$ then $b$ must equal $c$. But here $(pi alpha)(pi alpha) = 1$, while $(pi alpha)(alpha pi) = (pi alpha)(alpha^{-1}pi^{-1}) = 1$. So indeed, $pi alpha = alpha pi$.] This implies that $H doteq {1,alpha, pi, pialpha }$ is closed under composition, and as every element in $H$ has its inverse in $H$, it follows that $H$ is a subgroup, and has order 4.






        share|cite|improve this answer











        $endgroup$



        Well, you can show that $G$ has a subgroup of order 4 directly. If $G$ has an element of order 4 or 8 then we are already done [why?]



        Otherwise let $pi$ and $alpha$ be two distinct elements in $G$ of order 2. So $pi = pi^{-1}$ and $alpha = alpha^{-1}$. Then $pialpha$ must have order 2 as well, lest there be an element of order 4 or 8 which would imply that we are done. This implies that $pi alpha = alpha pi$. [Indeed, for any elements $a,b,c$ in any group $G'$, if $ab=ac$ then $b$ must equal $c$. But here $(pi alpha)(pi alpha) = 1$, while $(pi alpha)(alpha pi) = (pi alpha)(alpha^{-1}pi^{-1}) = 1$. So indeed, $pi alpha = alpha pi$.] This implies that $H doteq {1,alpha, pi, pialpha }$ is closed under composition, and as every element in $H$ has its inverse in $H$, it follows that $H$ is a subgroup, and has order 4.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 15 at 17:43

























        answered Feb 15 at 17:35









        MikeMike

        4,171412




        4,171412























            6












            $begingroup$

            If $G$ has no subgroup of order $4$, then every $gne e$ has order $2$. Then for any $g,h$,
            $$
            e=(gh)^2=ghgh=ghg^{-1}h^{-1},
            $$            i.e. $G$ is abelian. Then ${e,g,h,gh}$ would be a subgroup of order $4$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              why every $gneq e$ has order $2$ if $G$ does not have subgroup of order $4$? Also why $ghgh=ghg^{-1}h^{-1}$?
              $endgroup$
              – Baduk
              Feb 15 at 18:21












            • $begingroup$
              @Baduk: Apply Lagrange's Theorem. Also, we have $x^{-1}=x$ iff $x^2=e$. (Why?)
              $endgroup$
              – Shaun
              Feb 15 at 18:50


















            6












            $begingroup$

            If $G$ has no subgroup of order $4$, then every $gne e$ has order $2$. Then for any $g,h$,
            $$
            e=(gh)^2=ghgh=ghg^{-1}h^{-1},
            $$            i.e. $G$ is abelian. Then ${e,g,h,gh}$ would be a subgroup of order $4$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              why every $gneq e$ has order $2$ if $G$ does not have subgroup of order $4$? Also why $ghgh=ghg^{-1}h^{-1}$?
              $endgroup$
              – Baduk
              Feb 15 at 18:21












            • $begingroup$
              @Baduk: Apply Lagrange's Theorem. Also, we have $x^{-1}=x$ iff $x^2=e$. (Why?)
              $endgroup$
              – Shaun
              Feb 15 at 18:50
















            6












            6








            6





            $begingroup$

            If $G$ has no subgroup of order $4$, then every $gne e$ has order $2$. Then for any $g,h$,
            $$
            e=(gh)^2=ghgh=ghg^{-1}h^{-1},
            $$            i.e. $G$ is abelian. Then ${e,g,h,gh}$ would be a subgroup of order $4$.






            share|cite|improve this answer









            $endgroup$



            If $G$ has no subgroup of order $4$, then every $gne e$ has order $2$. Then for any $g,h$,
            $$
            e=(gh)^2=ghgh=ghg^{-1}h^{-1},
            $$            i.e. $G$ is abelian. Then ${e,g,h,gh}$ would be a subgroup of order $4$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 15 at 17:33









            SongSong

            14.4k1635




            14.4k1635












            • $begingroup$
              why every $gneq e$ has order $2$ if $G$ does not have subgroup of order $4$? Also why $ghgh=ghg^{-1}h^{-1}$?
              $endgroup$
              – Baduk
              Feb 15 at 18:21












            • $begingroup$
              @Baduk: Apply Lagrange's Theorem. Also, we have $x^{-1}=x$ iff $x^2=e$. (Why?)
              $endgroup$
              – Shaun
              Feb 15 at 18:50




















            • $begingroup$
              why every $gneq e$ has order $2$ if $G$ does not have subgroup of order $4$? Also why $ghgh=ghg^{-1}h^{-1}$?
              $endgroup$
              – Baduk
              Feb 15 at 18:21












            • $begingroup$
              @Baduk: Apply Lagrange's Theorem. Also, we have $x^{-1}=x$ iff $x^2=e$. (Why?)
              $endgroup$
              – Shaun
              Feb 15 at 18:50


















            $begingroup$
            why every $gneq e$ has order $2$ if $G$ does not have subgroup of order $4$? Also why $ghgh=ghg^{-1}h^{-1}$?
            $endgroup$
            – Baduk
            Feb 15 at 18:21






            $begingroup$
            why every $gneq e$ has order $2$ if $G$ does not have subgroup of order $4$? Also why $ghgh=ghg^{-1}h^{-1}$?
            $endgroup$
            – Baduk
            Feb 15 at 18:21














            $begingroup$
            @Baduk: Apply Lagrange's Theorem. Also, we have $x^{-1}=x$ iff $x^2=e$. (Why?)
            $endgroup$
            – Shaun
            Feb 15 at 18:50






            $begingroup$
            @Baduk: Apply Lagrange's Theorem. Also, we have $x^{-1}=x$ iff $x^2=e$. (Why?)
            $endgroup$
            – Shaun
            Feb 15 at 18:50













            3












            $begingroup$

            Hint: No group has exactly two elements of order two.




            Assume no subgroup of order four exists. Apply Lagrange's Theorem. Apply the hint. See @Mike's answer.




            Reference for the lemma that is the hint:



            Exercise 4.61 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)"






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
              $endgroup$
              – Gibbs
              Feb 15 at 17:58






            • 1




              $begingroup$
              @Gibbs, I've included a reference.
              $endgroup$
              – Shaun
              Feb 15 at 18:01
















            3












            $begingroup$

            Hint: No group has exactly two elements of order two.




            Assume no subgroup of order four exists. Apply Lagrange's Theorem. Apply the hint. See @Mike's answer.




            Reference for the lemma that is the hint:



            Exercise 4.61 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)"






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
              $endgroup$
              – Gibbs
              Feb 15 at 17:58






            • 1




              $begingroup$
              @Gibbs, I've included a reference.
              $endgroup$
              – Shaun
              Feb 15 at 18:01














            3












            3








            3





            $begingroup$

            Hint: No group has exactly two elements of order two.




            Assume no subgroup of order four exists. Apply Lagrange's Theorem. Apply the hint. See @Mike's answer.




            Reference for the lemma that is the hint:



            Exercise 4.61 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)"






            share|cite|improve this answer











            $endgroup$



            Hint: No group has exactly two elements of order two.




            Assume no subgroup of order four exists. Apply Lagrange's Theorem. Apply the hint. See @Mike's answer.




            Reference for the lemma that is the hint:



            Exercise 4.61 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)"







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 15 at 18:09

























            answered Feb 15 at 17:31









            ShaunShaun

            9,268113684




            9,268113684












            • $begingroup$
              While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
              $endgroup$
              – Gibbs
              Feb 15 at 17:58






            • 1




              $begingroup$
              @Gibbs, I've included a reference.
              $endgroup$
              – Shaun
              Feb 15 at 18:01


















            • $begingroup$
              While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
              $endgroup$
              – Gibbs
              Feb 15 at 17:58






            • 1




              $begingroup$
              @Gibbs, I've included a reference.
              $endgroup$
              – Shaun
              Feb 15 at 18:01
















            $begingroup$
            While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
            $endgroup$
            – Gibbs
            Feb 15 at 17:58




            $begingroup$
            While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review
            $endgroup$
            – Gibbs
            Feb 15 at 17:58




            1




            1




            $begingroup$
            @Gibbs, I've included a reference.
            $endgroup$
            – Shaun
            Feb 15 at 18:01




            $begingroup$
            @Gibbs, I've included a reference.
            $endgroup$
            – Shaun
            Feb 15 at 18:01


















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