Construct a nonabelian group of order 44 Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)The dihedral group $Vlanglealpharangle$Number of Sylow bases of a certain group of order 60Nonabelian group of order $p^3$ for odd prime $p$ and exponent $p$The only group of order $255$ is $mathbb Z_255$ ( Using Sylow and the $N/C$ Theorem)Number of elements of order $11$ in group of order $1331$Classifying groups of order $20$Construct a non-abelian group of order 75order of automorphism group of an abelian group of order 75Understanding a group of order $2^25.97^2$Understanding semidirect product for group of order 30

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Construct a nonabelian group of order 44



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)The dihedral group $Vlanglealpharangle$Number of Sylow bases of a certain group of order 60Nonabelian group of order $p^3$ for odd prime $p$ and exponent $p$The only group of order $255$ is $mathbb Z_255$ ( Using Sylow and the $N/C$ Theorem)Number of elements of order $11$ in group of order $1331$Classifying groups of order $20$Construct a non-abelian group of order 75order of automorphism group of an abelian group of order 75Understanding a group of order $2^25.97^2$Understanding semidirect product for group of order 30










6












$begingroup$


Let $G$ be a group s.t. $|G|=44=2^211$. Using Sylow's Theorems, I have deduced that there is a unique Sylow $11$-subgroup of $G$; we shall call it $R$. Let $P$ be a Sylow $2$-subgroup of $G$. Then we have $G=Prtimes R$ and a homomorphism



$$gamma: P rightarrow Aut(R)=Aut(mathbbZ_11)cong(mathbbZ_10,+) .$$



Is this all correct so far?



So what about $gamma(p)=phi_p$ where $phi_p(r)=r^5$. I thought this because $tilde5inmathbbZ_10$ has order $4$ so the order of any element of $P$ could divide it... or something...



So I was thinking the group would be something like



$$G= langle p,r | p^4=r^11 prp^-1=r^5 rangle .$$



Any insight is greatly appreciated! Thanks! I would like to know both where I went wrong and how to do it correctly.




Did I do the above right? Identifying $mathbbZ_11$ with the additive group of $mathbbZ_10$? Or should I look at it multiplicatively, because I don't understand how that isomorphism works so it doesn't make sense to define the conjugation that makes the semi-direct product well defined based on elements of the additive group $mathbbZ_10$, but instead realize that $10 in U(mathbbZ_11)$ has order $2$ so we can have a group presentation something like:



$G = langle p, r | p^2=r^11=1 , prp^-1=r^10 rangle$



Insight appreciated!



I understand the dihedral group of the $22$-gon works now, thank you. Can somebody help me with my approach in constructing a non-abelian group of order $44$ via the methods I've been using? Thanks!










share|cite|improve this question











$endgroup$











  • $begingroup$
    I think you meant $r^11$ (r^11), not $r^11$ (r^11)
    $endgroup$
    – J. W. Tanner
    4 hours ago










  • $begingroup$
    I've taken the liberty of apply the correction J.W. Tanner mentioned, as well as a few other minor fixes.
    $endgroup$
    – Travis
    3 hours ago










  • $begingroup$
    I don’t understand the words, “because $tilde5inBbb Z_10$ has order $4$”
    $endgroup$
    – Lubin
    2 hours ago










  • $begingroup$
    Doesn't $tilde5 in mathbbZ_10$ have order $2$?
    $endgroup$
    – Peter Shor
    2 hours ago










  • $begingroup$
    And if you have a non-abelian group of order 22, isn't it easy to find one of order 44?
    $endgroup$
    – Peter Shor
    2 hours ago















6












$begingroup$


Let $G$ be a group s.t. $|G|=44=2^211$. Using Sylow's Theorems, I have deduced that there is a unique Sylow $11$-subgroup of $G$; we shall call it $R$. Let $P$ be a Sylow $2$-subgroup of $G$. Then we have $G=Prtimes R$ and a homomorphism



$$gamma: P rightarrow Aut(R)=Aut(mathbbZ_11)cong(mathbbZ_10,+) .$$



Is this all correct so far?



So what about $gamma(p)=phi_p$ where $phi_p(r)=r^5$. I thought this because $tilde5inmathbbZ_10$ has order $4$ so the order of any element of $P$ could divide it... or something...



So I was thinking the group would be something like



$$G= langle p,r | p^4=r^11 prp^-1=r^5 rangle .$$



Any insight is greatly appreciated! Thanks! I would like to know both where I went wrong and how to do it correctly.




Did I do the above right? Identifying $mathbbZ_11$ with the additive group of $mathbbZ_10$? Or should I look at it multiplicatively, because I don't understand how that isomorphism works so it doesn't make sense to define the conjugation that makes the semi-direct product well defined based on elements of the additive group $mathbbZ_10$, but instead realize that $10 in U(mathbbZ_11)$ has order $2$ so we can have a group presentation something like:



$G = langle p, r | p^2=r^11=1 , prp^-1=r^10 rangle$



Insight appreciated!



I understand the dihedral group of the $22$-gon works now, thank you. Can somebody help me with my approach in constructing a non-abelian group of order $44$ via the methods I've been using? Thanks!










share|cite|improve this question











$endgroup$











  • $begingroup$
    I think you meant $r^11$ (r^11), not $r^11$ (r^11)
    $endgroup$
    – J. W. Tanner
    4 hours ago










  • $begingroup$
    I've taken the liberty of apply the correction J.W. Tanner mentioned, as well as a few other minor fixes.
    $endgroup$
    – Travis
    3 hours ago










  • $begingroup$
    I don’t understand the words, “because $tilde5inBbb Z_10$ has order $4$”
    $endgroup$
    – Lubin
    2 hours ago










  • $begingroup$
    Doesn't $tilde5 in mathbbZ_10$ have order $2$?
    $endgroup$
    – Peter Shor
    2 hours ago










  • $begingroup$
    And if you have a non-abelian group of order 22, isn't it easy to find one of order 44?
    $endgroup$
    – Peter Shor
    2 hours ago













6












6








6


2



$begingroup$


Let $G$ be a group s.t. $|G|=44=2^211$. Using Sylow's Theorems, I have deduced that there is a unique Sylow $11$-subgroup of $G$; we shall call it $R$. Let $P$ be a Sylow $2$-subgroup of $G$. Then we have $G=Prtimes R$ and a homomorphism



$$gamma: P rightarrow Aut(R)=Aut(mathbbZ_11)cong(mathbbZ_10,+) .$$



Is this all correct so far?



So what about $gamma(p)=phi_p$ where $phi_p(r)=r^5$. I thought this because $tilde5inmathbbZ_10$ has order $4$ so the order of any element of $P$ could divide it... or something...



So I was thinking the group would be something like



$$G= langle p,r | p^4=r^11 prp^-1=r^5 rangle .$$



Any insight is greatly appreciated! Thanks! I would like to know both where I went wrong and how to do it correctly.




Did I do the above right? Identifying $mathbbZ_11$ with the additive group of $mathbbZ_10$? Or should I look at it multiplicatively, because I don't understand how that isomorphism works so it doesn't make sense to define the conjugation that makes the semi-direct product well defined based on elements of the additive group $mathbbZ_10$, but instead realize that $10 in U(mathbbZ_11)$ has order $2$ so we can have a group presentation something like:



$G = langle p, r | p^2=r^11=1 , prp^-1=r^10 rangle$



Insight appreciated!



I understand the dihedral group of the $22$-gon works now, thank you. Can somebody help me with my approach in constructing a non-abelian group of order $44$ via the methods I've been using? Thanks!










share|cite|improve this question











$endgroup$




Let $G$ be a group s.t. $|G|=44=2^211$. Using Sylow's Theorems, I have deduced that there is a unique Sylow $11$-subgroup of $G$; we shall call it $R$. Let $P$ be a Sylow $2$-subgroup of $G$. Then we have $G=Prtimes R$ and a homomorphism



$$gamma: P rightarrow Aut(R)=Aut(mathbbZ_11)cong(mathbbZ_10,+) .$$



Is this all correct so far?



So what about $gamma(p)=phi_p$ where $phi_p(r)=r^5$. I thought this because $tilde5inmathbbZ_10$ has order $4$ so the order of any element of $P$ could divide it... or something...



So I was thinking the group would be something like



$$G= langle p,r | p^4=r^11 prp^-1=r^5 rangle .$$



Any insight is greatly appreciated! Thanks! I would like to know both where I went wrong and how to do it correctly.




Did I do the above right? Identifying $mathbbZ_11$ with the additive group of $mathbbZ_10$? Or should I look at it multiplicatively, because I don't understand how that isomorphism works so it doesn't make sense to define the conjugation that makes the semi-direct product well defined based on elements of the additive group $mathbbZ_10$, but instead realize that $10 in U(mathbbZ_11)$ has order $2$ so we can have a group presentation something like:



$G = langle p, r | p^2=r^11=1 , prp^-1=r^10 rangle$



Insight appreciated!



I understand the dihedral group of the $22$-gon works now, thank you. Can somebody help me with my approach in constructing a non-abelian group of order $44$ via the methods I've been using? Thanks!







abstract-algebra group-theory sylow-theory group-presentation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Travis

64.6k769152




64.6k769152










asked 5 hours ago









Mathematical MushroomMathematical Mushroom

22418




22418











  • $begingroup$
    I think you meant $r^11$ (r^11), not $r^11$ (r^11)
    $endgroup$
    – J. W. Tanner
    4 hours ago










  • $begingroup$
    I've taken the liberty of apply the correction J.W. Tanner mentioned, as well as a few other minor fixes.
    $endgroup$
    – Travis
    3 hours ago










  • $begingroup$
    I don’t understand the words, “because $tilde5inBbb Z_10$ has order $4$”
    $endgroup$
    – Lubin
    2 hours ago










  • $begingroup$
    Doesn't $tilde5 in mathbbZ_10$ have order $2$?
    $endgroup$
    – Peter Shor
    2 hours ago










  • $begingroup$
    And if you have a non-abelian group of order 22, isn't it easy to find one of order 44?
    $endgroup$
    – Peter Shor
    2 hours ago
















  • $begingroup$
    I think you meant $r^11$ (r^11), not $r^11$ (r^11)
    $endgroup$
    – J. W. Tanner
    4 hours ago










  • $begingroup$
    I've taken the liberty of apply the correction J.W. Tanner mentioned, as well as a few other minor fixes.
    $endgroup$
    – Travis
    3 hours ago










  • $begingroup$
    I don’t understand the words, “because $tilde5inBbb Z_10$ has order $4$”
    $endgroup$
    – Lubin
    2 hours ago










  • $begingroup$
    Doesn't $tilde5 in mathbbZ_10$ have order $2$?
    $endgroup$
    – Peter Shor
    2 hours ago










  • $begingroup$
    And if you have a non-abelian group of order 22, isn't it easy to find one of order 44?
    $endgroup$
    – Peter Shor
    2 hours ago















$begingroup$
I think you meant $r^11$ (r^11), not $r^11$ (r^11)
$endgroup$
– J. W. Tanner
4 hours ago




$begingroup$
I think you meant $r^11$ (r^11), not $r^11$ (r^11)
$endgroup$
– J. W. Tanner
4 hours ago












$begingroup$
I've taken the liberty of apply the correction J.W. Tanner mentioned, as well as a few other minor fixes.
$endgroup$
– Travis
3 hours ago




$begingroup$
I've taken the liberty of apply the correction J.W. Tanner mentioned, as well as a few other minor fixes.
$endgroup$
– Travis
3 hours ago












$begingroup$
I don’t understand the words, “because $tilde5inBbb Z_10$ has order $4$”
$endgroup$
– Lubin
2 hours ago




$begingroup$
I don’t understand the words, “because $tilde5inBbb Z_10$ has order $4$”
$endgroup$
– Lubin
2 hours ago












$begingroup$
Doesn't $tilde5 in mathbbZ_10$ have order $2$?
$endgroup$
– Peter Shor
2 hours ago




$begingroup$
Doesn't $tilde5 in mathbbZ_10$ have order $2$?
$endgroup$
– Peter Shor
2 hours ago












$begingroup$
And if you have a non-abelian group of order 22, isn't it easy to find one of order 44?
$endgroup$
– Peter Shor
2 hours ago




$begingroup$
And if you have a non-abelian group of order 22, isn't it easy to find one of order 44?
$endgroup$
– Peter Shor
2 hours ago










2 Answers
2






active

oldest

votes


















1












$begingroup$

No element of $mathbb Z_10$ has order four (why not?) and there is one element of order 2 ($5inmathbb Z_10$ under addition, $10cong -1in mathbb Z_11^x$ under multiplication.), so our possibilities are quite limited. There are two groups of order $4$, and either will work as our $R$. To give a nonabelian group, we need to pick a nontrivial homomorphism, as you pointed out.



So at least one generator of $R$ has to map to our order-two element. In the case of the Klein four group, there appear to be three possibilities, but I claim that up to an isomorphism of the Klein four group, there is only one possibility.



Actually, this exhausts the possibilities for groups of order 44: we have two abelian groups, $mathbb Z_11timesmathbbZ_4$, $mathbb Z_11times mathbbZ_2times mathbb Z_2$, and two nonabelian groups: $mathbbZ_11 rtimes mathbb Z_4 = langle a,b mid a^11, b^4, b^-1ab = a^-1rangle$ and $mathbb Z_11rtimes(mathbb Z_2 times mathbb Z_2) = langle a, b, c mid a^11, b^2, c^2, [b,c], b^-1ab = c^-1ac = a^-1 rangle$. I think the latter is $D_22$ (for 22-gon, not order of group), while the former has an element of order $4$.






share|cite|improve this answer











$endgroup$




















    0












    $begingroup$

    You're on the right track (but NB your semidirect product is written in the wrong order). To analyze the possible maps $gamma : P to operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$, we consider separately the cases $P cong Bbb Z_2 times Bbb Z_2$ and $P cong Bbb Z_4$.



    In the case $P cong Bbb Z_4$, $P$ is generated by a single element, $[1]$, of order $4$, and so $$operatornameid_Bbb Z_11 = gamma([0]) = gamma([1] + [1] + [1] + [1]) = gamma([1])^4 .$$ So, $gamma([1])$ has order dividing $4$, and the only such elements of $operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$ are $operatornameid_Bbb Z_11 leftrightarrow [0]$ and $(x mapsto -x) leftrightarrow [5]$.



    • If $gamma([1]) = operatornameid_Bbb Z_11$, then $gamma$ is the trivial homomorphism $Bbb Z_4 to operatornameAut(Bbb Z_11)$. This gives the direct product $G cong Bbb Z_11 times Bbb Z_4 cong Bbb Z_44 .$


    • If $gamma([1]) = (x mapsto -x)$, then $gamma([b])([c]) = (-1)^b [c]$, and the semidirect product $G = Bbb Z_11 rtimes_gamma Bbb Z_4$ is defined by
      $$([a], [b]) cdot ([c], [d]) = ([a] + (-1)^b [c], [b] + [d]) .$$ It's apparent from the multiplication rule that this group is nonabelian. The fact that the group is generated by $u := ([1], [0])$ and $v := ([0], [1])$ can be used to construct an explicit presentation of this group and to show that $G$ is the dicyclic group of order $44$.


    One can analyze the case $P cong Bbb Z_2 times Bbb Z_2$ similarly, and this case gives rise to two more groups up to isomorphism, namely the abelian group $Bbb Z_11 times Bbb Z_2 times Bbb Z_2$ and the (nonabelian) dihedral group $D_44 cong D_22 times Bbb Z_2$.






    share|cite|improve this answer











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      2 Answers
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      2 Answers
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      active

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      1












      $begingroup$

      No element of $mathbb Z_10$ has order four (why not?) and there is one element of order 2 ($5inmathbb Z_10$ under addition, $10cong -1in mathbb Z_11^x$ under multiplication.), so our possibilities are quite limited. There are two groups of order $4$, and either will work as our $R$. To give a nonabelian group, we need to pick a nontrivial homomorphism, as you pointed out.



      So at least one generator of $R$ has to map to our order-two element. In the case of the Klein four group, there appear to be three possibilities, but I claim that up to an isomorphism of the Klein four group, there is only one possibility.



      Actually, this exhausts the possibilities for groups of order 44: we have two abelian groups, $mathbb Z_11timesmathbbZ_4$, $mathbb Z_11times mathbbZ_2times mathbb Z_2$, and two nonabelian groups: $mathbbZ_11 rtimes mathbb Z_4 = langle a,b mid a^11, b^4, b^-1ab = a^-1rangle$ and $mathbb Z_11rtimes(mathbb Z_2 times mathbb Z_2) = langle a, b, c mid a^11, b^2, c^2, [b,c], b^-1ab = c^-1ac = a^-1 rangle$. I think the latter is $D_22$ (for 22-gon, not order of group), while the former has an element of order $4$.






      share|cite|improve this answer











      $endgroup$

















        1












        $begingroup$

        No element of $mathbb Z_10$ has order four (why not?) and there is one element of order 2 ($5inmathbb Z_10$ under addition, $10cong -1in mathbb Z_11^x$ under multiplication.), so our possibilities are quite limited. There are two groups of order $4$, and either will work as our $R$. To give a nonabelian group, we need to pick a nontrivial homomorphism, as you pointed out.



        So at least one generator of $R$ has to map to our order-two element. In the case of the Klein four group, there appear to be three possibilities, but I claim that up to an isomorphism of the Klein four group, there is only one possibility.



        Actually, this exhausts the possibilities for groups of order 44: we have two abelian groups, $mathbb Z_11timesmathbbZ_4$, $mathbb Z_11times mathbbZ_2times mathbb Z_2$, and two nonabelian groups: $mathbbZ_11 rtimes mathbb Z_4 = langle a,b mid a^11, b^4, b^-1ab = a^-1rangle$ and $mathbb Z_11rtimes(mathbb Z_2 times mathbb Z_2) = langle a, b, c mid a^11, b^2, c^2, [b,c], b^-1ab = c^-1ac = a^-1 rangle$. I think the latter is $D_22$ (for 22-gon, not order of group), while the former has an element of order $4$.






        share|cite|improve this answer











        $endgroup$















          1












          1








          1





          $begingroup$

          No element of $mathbb Z_10$ has order four (why not?) and there is one element of order 2 ($5inmathbb Z_10$ under addition, $10cong -1in mathbb Z_11^x$ under multiplication.), so our possibilities are quite limited. There are two groups of order $4$, and either will work as our $R$. To give a nonabelian group, we need to pick a nontrivial homomorphism, as you pointed out.



          So at least one generator of $R$ has to map to our order-two element. In the case of the Klein four group, there appear to be three possibilities, but I claim that up to an isomorphism of the Klein four group, there is only one possibility.



          Actually, this exhausts the possibilities for groups of order 44: we have two abelian groups, $mathbb Z_11timesmathbbZ_4$, $mathbb Z_11times mathbbZ_2times mathbb Z_2$, and two nonabelian groups: $mathbbZ_11 rtimes mathbb Z_4 = langle a,b mid a^11, b^4, b^-1ab = a^-1rangle$ and $mathbb Z_11rtimes(mathbb Z_2 times mathbb Z_2) = langle a, b, c mid a^11, b^2, c^2, [b,c], b^-1ab = c^-1ac = a^-1 rangle$. I think the latter is $D_22$ (for 22-gon, not order of group), while the former has an element of order $4$.






          share|cite|improve this answer











          $endgroup$



          No element of $mathbb Z_10$ has order four (why not?) and there is one element of order 2 ($5inmathbb Z_10$ under addition, $10cong -1in mathbb Z_11^x$ under multiplication.), so our possibilities are quite limited. There are two groups of order $4$, and either will work as our $R$. To give a nonabelian group, we need to pick a nontrivial homomorphism, as you pointed out.



          So at least one generator of $R$ has to map to our order-two element. In the case of the Klein four group, there appear to be three possibilities, but I claim that up to an isomorphism of the Klein four group, there is only one possibility.



          Actually, this exhausts the possibilities for groups of order 44: we have two abelian groups, $mathbb Z_11timesmathbbZ_4$, $mathbb Z_11times mathbbZ_2times mathbb Z_2$, and two nonabelian groups: $mathbbZ_11 rtimes mathbb Z_4 = langle a,b mid a^11, b^4, b^-1ab = a^-1rangle$ and $mathbb Z_11rtimes(mathbb Z_2 times mathbb Z_2) = langle a, b, c mid a^11, b^2, c^2, [b,c], b^-1ab = c^-1ac = a^-1 rangle$. I think the latter is $D_22$ (for 22-gon, not order of group), while the former has an element of order $4$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          Rylee LymanRylee Lyman

          646211




          646211





















              0












              $begingroup$

              You're on the right track (but NB your semidirect product is written in the wrong order). To analyze the possible maps $gamma : P to operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$, we consider separately the cases $P cong Bbb Z_2 times Bbb Z_2$ and $P cong Bbb Z_4$.



              In the case $P cong Bbb Z_4$, $P$ is generated by a single element, $[1]$, of order $4$, and so $$operatornameid_Bbb Z_11 = gamma([0]) = gamma([1] + [1] + [1] + [1]) = gamma([1])^4 .$$ So, $gamma([1])$ has order dividing $4$, and the only such elements of $operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$ are $operatornameid_Bbb Z_11 leftrightarrow [0]$ and $(x mapsto -x) leftrightarrow [5]$.



              • If $gamma([1]) = operatornameid_Bbb Z_11$, then $gamma$ is the trivial homomorphism $Bbb Z_4 to operatornameAut(Bbb Z_11)$. This gives the direct product $G cong Bbb Z_11 times Bbb Z_4 cong Bbb Z_44 .$


              • If $gamma([1]) = (x mapsto -x)$, then $gamma([b])([c]) = (-1)^b [c]$, and the semidirect product $G = Bbb Z_11 rtimes_gamma Bbb Z_4$ is defined by
                $$([a], [b]) cdot ([c], [d]) = ([a] + (-1)^b [c], [b] + [d]) .$$ It's apparent from the multiplication rule that this group is nonabelian. The fact that the group is generated by $u := ([1], [0])$ and $v := ([0], [1])$ can be used to construct an explicit presentation of this group and to show that $G$ is the dicyclic group of order $44$.


              One can analyze the case $P cong Bbb Z_2 times Bbb Z_2$ similarly, and this case gives rise to two more groups up to isomorphism, namely the abelian group $Bbb Z_11 times Bbb Z_2 times Bbb Z_2$ and the (nonabelian) dihedral group $D_44 cong D_22 times Bbb Z_2$.






              share|cite|improve this answer











              $endgroup$

















                0












                $begingroup$

                You're on the right track (but NB your semidirect product is written in the wrong order). To analyze the possible maps $gamma : P to operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$, we consider separately the cases $P cong Bbb Z_2 times Bbb Z_2$ and $P cong Bbb Z_4$.



                In the case $P cong Bbb Z_4$, $P$ is generated by a single element, $[1]$, of order $4$, and so $$operatornameid_Bbb Z_11 = gamma([0]) = gamma([1] + [1] + [1] + [1]) = gamma([1])^4 .$$ So, $gamma([1])$ has order dividing $4$, and the only such elements of $operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$ are $operatornameid_Bbb Z_11 leftrightarrow [0]$ and $(x mapsto -x) leftrightarrow [5]$.



                • If $gamma([1]) = operatornameid_Bbb Z_11$, then $gamma$ is the trivial homomorphism $Bbb Z_4 to operatornameAut(Bbb Z_11)$. This gives the direct product $G cong Bbb Z_11 times Bbb Z_4 cong Bbb Z_44 .$


                • If $gamma([1]) = (x mapsto -x)$, then $gamma([b])([c]) = (-1)^b [c]$, and the semidirect product $G = Bbb Z_11 rtimes_gamma Bbb Z_4$ is defined by
                  $$([a], [b]) cdot ([c], [d]) = ([a] + (-1)^b [c], [b] + [d]) .$$ It's apparent from the multiplication rule that this group is nonabelian. The fact that the group is generated by $u := ([1], [0])$ and $v := ([0], [1])$ can be used to construct an explicit presentation of this group and to show that $G$ is the dicyclic group of order $44$.


                One can analyze the case $P cong Bbb Z_2 times Bbb Z_2$ similarly, and this case gives rise to two more groups up to isomorphism, namely the abelian group $Bbb Z_11 times Bbb Z_2 times Bbb Z_2$ and the (nonabelian) dihedral group $D_44 cong D_22 times Bbb Z_2$.






                share|cite|improve this answer











                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  You're on the right track (but NB your semidirect product is written in the wrong order). To analyze the possible maps $gamma : P to operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$, we consider separately the cases $P cong Bbb Z_2 times Bbb Z_2$ and $P cong Bbb Z_4$.



                  In the case $P cong Bbb Z_4$, $P$ is generated by a single element, $[1]$, of order $4$, and so $$operatornameid_Bbb Z_11 = gamma([0]) = gamma([1] + [1] + [1] + [1]) = gamma([1])^4 .$$ So, $gamma([1])$ has order dividing $4$, and the only such elements of $operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$ are $operatornameid_Bbb Z_11 leftrightarrow [0]$ and $(x mapsto -x) leftrightarrow [5]$.



                  • If $gamma([1]) = operatornameid_Bbb Z_11$, then $gamma$ is the trivial homomorphism $Bbb Z_4 to operatornameAut(Bbb Z_11)$. This gives the direct product $G cong Bbb Z_11 times Bbb Z_4 cong Bbb Z_44 .$


                  • If $gamma([1]) = (x mapsto -x)$, then $gamma([b])([c]) = (-1)^b [c]$, and the semidirect product $G = Bbb Z_11 rtimes_gamma Bbb Z_4$ is defined by
                    $$([a], [b]) cdot ([c], [d]) = ([a] + (-1)^b [c], [b] + [d]) .$$ It's apparent from the multiplication rule that this group is nonabelian. The fact that the group is generated by $u := ([1], [0])$ and $v := ([0], [1])$ can be used to construct an explicit presentation of this group and to show that $G$ is the dicyclic group of order $44$.


                  One can analyze the case $P cong Bbb Z_2 times Bbb Z_2$ similarly, and this case gives rise to two more groups up to isomorphism, namely the abelian group $Bbb Z_11 times Bbb Z_2 times Bbb Z_2$ and the (nonabelian) dihedral group $D_44 cong D_22 times Bbb Z_2$.






                  share|cite|improve this answer











                  $endgroup$



                  You're on the right track (but NB your semidirect product is written in the wrong order). To analyze the possible maps $gamma : P to operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$, we consider separately the cases $P cong Bbb Z_2 times Bbb Z_2$ and $P cong Bbb Z_4$.



                  In the case $P cong Bbb Z_4$, $P$ is generated by a single element, $[1]$, of order $4$, and so $$operatornameid_Bbb Z_11 = gamma([0]) = gamma([1] + [1] + [1] + [1]) = gamma([1])^4 .$$ So, $gamma([1])$ has order dividing $4$, and the only such elements of $operatornameAut(Bbb Z_11) cong (Bbb Z_10, +)$ are $operatornameid_Bbb Z_11 leftrightarrow [0]$ and $(x mapsto -x) leftrightarrow [5]$.



                  • If $gamma([1]) = operatornameid_Bbb Z_11$, then $gamma$ is the trivial homomorphism $Bbb Z_4 to operatornameAut(Bbb Z_11)$. This gives the direct product $G cong Bbb Z_11 times Bbb Z_4 cong Bbb Z_44 .$


                  • If $gamma([1]) = (x mapsto -x)$, then $gamma([b])([c]) = (-1)^b [c]$, and the semidirect product $G = Bbb Z_11 rtimes_gamma Bbb Z_4$ is defined by
                    $$([a], [b]) cdot ([c], [d]) = ([a] + (-1)^b [c], [b] + [d]) .$$ It's apparent from the multiplication rule that this group is nonabelian. The fact that the group is generated by $u := ([1], [0])$ and $v := ([0], [1])$ can be used to construct an explicit presentation of this group and to show that $G$ is the dicyclic group of order $44$.


                  One can analyze the case $P cong Bbb Z_2 times Bbb Z_2$ similarly, and this case gives rise to two more groups up to isomorphism, namely the abelian group $Bbb Z_11 times Bbb Z_2 times Bbb Z_2$ and the (nonabelian) dihedral group $D_44 cong D_22 times Bbb Z_2$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 1 hour ago

























                  answered 1 hour ago









                  TravisTravis

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