Does there exist a group that is both a free product and a direct product of nontrivial groups?












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Do there exist such nontrivial groups $A$, $B$, $C$ and $D$, such that $A times B cong C ast D$?



I failed to construct any examples, so I decided to try to prove they do not exist by contradiction.



If such groups exist, then $C$ and $D$ are disjoint subgroups of $A times B$. Suppose $w in F[x, y] {e}$, where $F[x, y]$ is the free group with generators $x$ and $y$. Suppose $(a_c, b_c) in C$, $(a_d, b_d) in D$ and $h: F[x, y] rightarrow A times B$ is a homomorphism, that maps $x$ to $(a_c, b_c)$ and $y$ to $(a_d, b_d)$. Then, by definition of free product $h(w) neq e$. Thus, either $pi_A(h(w)) neq e$ or $pi_B(h(w)) neq e$, where $pi_A$ and $pi_B$ are projections on $A times B$ onto $A$ and $B$ respectively. Thus every group word is not an identity either for $A$ or for $B$ and that results in ${A, B}$ generating the variety of all groups. And here I am stuck, failing to determine anything else.



Or do the examples actually exist?










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    Related: mathoverflow.net/questions/253052/… and math.stackexchange.com/questions/78227/…
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    – Michael Burr
    6 hours ago


















4












$begingroup$


Do there exist such nontrivial groups $A$, $B$, $C$ and $D$, such that $A times B cong C ast D$?



I failed to construct any examples, so I decided to try to prove they do not exist by contradiction.



If such groups exist, then $C$ and $D$ are disjoint subgroups of $A times B$. Suppose $w in F[x, y] {e}$, where $F[x, y]$ is the free group with generators $x$ and $y$. Suppose $(a_c, b_c) in C$, $(a_d, b_d) in D$ and $h: F[x, y] rightarrow A times B$ is a homomorphism, that maps $x$ to $(a_c, b_c)$ and $y$ to $(a_d, b_d)$. Then, by definition of free product $h(w) neq e$. Thus, either $pi_A(h(w)) neq e$ or $pi_B(h(w)) neq e$, where $pi_A$ and $pi_B$ are projections on $A times B$ onto $A$ and $B$ respectively. Thus every group word is not an identity either for $A$ or for $B$ and that results in ${A, B}$ generating the variety of all groups. And here I am stuck, failing to determine anything else.



Or do the examples actually exist?










share|cite|improve this question











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  • 1




    $begingroup$
    Related: mathoverflow.net/questions/253052/… and math.stackexchange.com/questions/78227/…
    $endgroup$
    – Michael Burr
    6 hours ago
















4












4








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$begingroup$


Do there exist such nontrivial groups $A$, $B$, $C$ and $D$, such that $A times B cong C ast D$?



I failed to construct any examples, so I decided to try to prove they do not exist by contradiction.



If such groups exist, then $C$ and $D$ are disjoint subgroups of $A times B$. Suppose $w in F[x, y] {e}$, where $F[x, y]$ is the free group with generators $x$ and $y$. Suppose $(a_c, b_c) in C$, $(a_d, b_d) in D$ and $h: F[x, y] rightarrow A times B$ is a homomorphism, that maps $x$ to $(a_c, b_c)$ and $y$ to $(a_d, b_d)$. Then, by definition of free product $h(w) neq e$. Thus, either $pi_A(h(w)) neq e$ or $pi_B(h(w)) neq e$, where $pi_A$ and $pi_B$ are projections on $A times B$ onto $A$ and $B$ respectively. Thus every group word is not an identity either for $A$ or for $B$ and that results in ${A, B}$ generating the variety of all groups. And here I am stuck, failing to determine anything else.



Or do the examples actually exist?










share|cite|improve this question











$endgroup$




Do there exist such nontrivial groups $A$, $B$, $C$ and $D$, such that $A times B cong C ast D$?



I failed to construct any examples, so I decided to try to prove they do not exist by contradiction.



If such groups exist, then $C$ and $D$ are disjoint subgroups of $A times B$. Suppose $w in F[x, y] {e}$, where $F[x, y]$ is the free group with generators $x$ and $y$. Suppose $(a_c, b_c) in C$, $(a_d, b_d) in D$ and $h: F[x, y] rightarrow A times B$ is a homomorphism, that maps $x$ to $(a_c, b_c)$ and $y$ to $(a_d, b_d)$. Then, by definition of free product $h(w) neq e$. Thus, either $pi_A(h(w)) neq e$ or $pi_B(h(w)) neq e$, where $pi_A$ and $pi_B$ are projections on $A times B$ onto $A$ and $B$ respectively. Thus every group word is not an identity either for $A$ or for $B$ and that results in ${A, B}$ generating the variety of all groups. And here I am stuck, failing to determine anything else.



Or do the examples actually exist?







abstract-algebra group-theory direct-product combinatorial-group-theory free-product






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edited 7 hours ago









Shaun

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asked 7 hours ago









Yanior WegYanior Weg

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1,94311140








  • 1




    $begingroup$
    Related: mathoverflow.net/questions/253052/… and math.stackexchange.com/questions/78227/…
    $endgroup$
    – Michael Burr
    6 hours ago
















  • 1




    $begingroup$
    Related: mathoverflow.net/questions/253052/… and math.stackexchange.com/questions/78227/…
    $endgroup$
    – Michael Burr
    6 hours ago










1




1




$begingroup$
Related: mathoverflow.net/questions/253052/… and math.stackexchange.com/questions/78227/…
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– Michael Burr
6 hours ago






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Related: mathoverflow.net/questions/253052/… and math.stackexchange.com/questions/78227/…
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– Michael Burr
6 hours ago












3 Answers
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Let me just replicate YCor's comment in the link that was given in the comments, with more details :




in a free product $Cast D$ with $C,D$ nontrivial, the intersection of any two nontrivial normal subgroups is nontrivial.




Indeed, let $H,K$ be two nontrivial normal subgroups of $C*D$. I'll assume for simplicity that $|C|,|D|$ are large enough so that for each $x,y$ there is a nontrivial $z$ with $z^{-1}neq x, zneq y$. For instance $|C|, |D|geq 4$ is good enough (take $x,y$, then there are at most two nontrivial $z$ such that $z=x$ or $z^{-1}=y$ : $x$ and $y^{-1}$; so if $|G|geq 4$, any nontrivial element different from $x$ and $y^{-1}$ works)



Let me say that $c_1d_1....c_nd_n$ is the reduced form of an element of $C*D$ if $c_iin C, d_jin D$, and the only $c_i,d_j$ allowed to be $1$ are $c_1$ and $d_n$. Clearly, if $x=c_1d_1....c_nd_n$ is the reduced form of $x$, and $ngeq 2$, then $xneq 1$ in $C*D$ ("clearly" here is to be understood as : it's a classical property of free products), moreover, if $n=1$ this is $1$ if and only if $c_1=d_1=1$.



Now let $x= c_1d_1...c_rd_r in H, y=c'_1d'_1...c'_sd'_s in K$ be nontrivial elements, with obvious notations, written in reduced form. The point will be that $[x,y]in Hcap K$ (this is obvious), and that, up to changing $x,y$ a bit, this can't be the trivial element.



Now up to conjugation by some element of $C$, one may assume $d_r = 1$ and $c_1neq 1$ (this can be done by the hypothesis on $|C|$ - and $c_r neq 1$, but that follows from the reduced form); and up to conjugation by some element of $D$, $c'_1 = 1$ , $d_s'neq 1$ (using the cardinality hypothesis on $D$ - and $d_1' neq 1$, but again this follows from the reduced form).



So with these hypotheses $[x,y] = c_1d_1...c_rd_1'....c_s'd_s'c_r^{-1}....d_1^{-1}c_1^{-1}d_s'^{-1}c_s'^{-1}...d_1'^{-1}$ which is written in reduced form, and is thus $neq 1$. Therefore $[x,y]in Hcap Ksetminus{1}$.



Apply this to your supposed $Atimes {1}, {1}times B$ to get a contradiction.



I don't know if there's an easier argument, or a not-too-complicated argument that encapsulates the low cardinality cases, but I guess for these you have to go "by hand" somehow; or perhaps you can adapt this argument to these cases by working a bit more. In any case I didn't want to bother with these cases, and this argument works in most cases and is pretty painless so in any case it's interesting to share






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    Let $f:Atimes Brightarrow C*D$ be an isomorphism where $A,B,C,D$ are non trivial groups. Let $a_0in A$ not trivial, for every $bin B$, $f(a_0)$ commutes with $f(b)$.



    Now we look the proof of the The Corollary 4.1.6 p. 187 of Magnus Karrass and Solitar.



    The first part of this proof asserts that if $f(a_0)$ is contained in the conjugated of a free factor, that is $f(a_0)$ is in $gCg^{-1}$ or is in $gDg^{-1}$ then so is $f(b)$ for every $bin B$. Without restricting the generality, we suppose that $f(a_0)$ and so $f(B)$ are contained in $gCg^{-1}$. Let $b_0$ a non trivial element of $B$ since $B$ is contained in the free conjugated factor $gCg^{-1}$ the same argument shows that $f(A)$ is contained in $gCg^{-1}$ this implies that $C*D$ is contained in a free conjugated factor. Contradiction.



    The second part of the proof shows that if $f(a_0)$ is not contained in a free conjugated factor, then there exists $u_c$ such $f(a_0)$ and $f(b)$ are a power of $u_c$.



    You can express $f(a_0)$ uniquely as a reduced sequence (Theorem 4.1.) this implies that there exists a unique element $u$ with minimal length such that $f(a_0)$ is a power of $u$ and if $f(a_0)$ is a power of $v$, $v$ is a power of $u$. This impllies that every element of $f(B)$ are power of $u$ and $f(B)$ and $B$ are cyclic. A similar argument shows that $f(A)$ and $A$ are cyclic, we deduce that $Atimes B$ is commutative. Contradicition since $C*D$ is not commutative.



    Reference.



    Combinatorial Group theory.



    Magnus, Karrass and Solitar.






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      To be a free product $Cast D$ implies the existence of an action on oriented tree such that $C$ is the stabilizer of some vertex $v_0$ and such that edge stabilizers are trivial (so that nontrivial element fix at most one vertex).



      Let $(a,b)$ be a nontrivial element of $C$; we can suppose that $aneq 1$ up to switch $A$ and $B$. Then in the Bass-Serre tree, $(a,b)$ fixes a unique vertex $v_0$, and hence this vertex is also fixed by the centralizer of $(a,b)$, and hence $(a,1)$ fixes $v_0$. Applying this to $(a,1)$ shows that $B$ fixes the vertex $v_0$. Choose $1neq b'in B$. It fixes the unique vertex $v_0$ and hence its centralizer fixes $v_0$, so $A$ fixes $v_0$. Finally $G=Atimes B$ fixes $v_0$, so $D=1$.






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        3 Answers
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        $begingroup$

        Let me just replicate YCor's comment in the link that was given in the comments, with more details :




        in a free product $Cast D$ with $C,D$ nontrivial, the intersection of any two nontrivial normal subgroups is nontrivial.




        Indeed, let $H,K$ be two nontrivial normal subgroups of $C*D$. I'll assume for simplicity that $|C|,|D|$ are large enough so that for each $x,y$ there is a nontrivial $z$ with $z^{-1}neq x, zneq y$. For instance $|C|, |D|geq 4$ is good enough (take $x,y$, then there are at most two nontrivial $z$ such that $z=x$ or $z^{-1}=y$ : $x$ and $y^{-1}$; so if $|G|geq 4$, any nontrivial element different from $x$ and $y^{-1}$ works)



        Let me say that $c_1d_1....c_nd_n$ is the reduced form of an element of $C*D$ if $c_iin C, d_jin D$, and the only $c_i,d_j$ allowed to be $1$ are $c_1$ and $d_n$. Clearly, if $x=c_1d_1....c_nd_n$ is the reduced form of $x$, and $ngeq 2$, then $xneq 1$ in $C*D$ ("clearly" here is to be understood as : it's a classical property of free products), moreover, if $n=1$ this is $1$ if and only if $c_1=d_1=1$.



        Now let $x= c_1d_1...c_rd_r in H, y=c'_1d'_1...c'_sd'_s in K$ be nontrivial elements, with obvious notations, written in reduced form. The point will be that $[x,y]in Hcap K$ (this is obvious), and that, up to changing $x,y$ a bit, this can't be the trivial element.



        Now up to conjugation by some element of $C$, one may assume $d_r = 1$ and $c_1neq 1$ (this can be done by the hypothesis on $|C|$ - and $c_r neq 1$, but that follows from the reduced form); and up to conjugation by some element of $D$, $c'_1 = 1$ , $d_s'neq 1$ (using the cardinality hypothesis on $D$ - and $d_1' neq 1$, but again this follows from the reduced form).



        So with these hypotheses $[x,y] = c_1d_1...c_rd_1'....c_s'd_s'c_r^{-1}....d_1^{-1}c_1^{-1}d_s'^{-1}c_s'^{-1}...d_1'^{-1}$ which is written in reduced form, and is thus $neq 1$. Therefore $[x,y]in Hcap Ksetminus{1}$.



        Apply this to your supposed $Atimes {1}, {1}times B$ to get a contradiction.



        I don't know if there's an easier argument, or a not-too-complicated argument that encapsulates the low cardinality cases, but I guess for these you have to go "by hand" somehow; or perhaps you can adapt this argument to these cases by working a bit more. In any case I didn't want to bother with these cases, and this argument works in most cases and is pretty painless so in any case it's interesting to share






        share|cite|improve this answer











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          3












          $begingroup$

          Let me just replicate YCor's comment in the link that was given in the comments, with more details :




          in a free product $Cast D$ with $C,D$ nontrivial, the intersection of any two nontrivial normal subgroups is nontrivial.




          Indeed, let $H,K$ be two nontrivial normal subgroups of $C*D$. I'll assume for simplicity that $|C|,|D|$ are large enough so that for each $x,y$ there is a nontrivial $z$ with $z^{-1}neq x, zneq y$. For instance $|C|, |D|geq 4$ is good enough (take $x,y$, then there are at most two nontrivial $z$ such that $z=x$ or $z^{-1}=y$ : $x$ and $y^{-1}$; so if $|G|geq 4$, any nontrivial element different from $x$ and $y^{-1}$ works)



          Let me say that $c_1d_1....c_nd_n$ is the reduced form of an element of $C*D$ if $c_iin C, d_jin D$, and the only $c_i,d_j$ allowed to be $1$ are $c_1$ and $d_n$. Clearly, if $x=c_1d_1....c_nd_n$ is the reduced form of $x$, and $ngeq 2$, then $xneq 1$ in $C*D$ ("clearly" here is to be understood as : it's a classical property of free products), moreover, if $n=1$ this is $1$ if and only if $c_1=d_1=1$.



          Now let $x= c_1d_1...c_rd_r in H, y=c'_1d'_1...c'_sd'_s in K$ be nontrivial elements, with obvious notations, written in reduced form. The point will be that $[x,y]in Hcap K$ (this is obvious), and that, up to changing $x,y$ a bit, this can't be the trivial element.



          Now up to conjugation by some element of $C$, one may assume $d_r = 1$ and $c_1neq 1$ (this can be done by the hypothesis on $|C|$ - and $c_r neq 1$, but that follows from the reduced form); and up to conjugation by some element of $D$, $c'_1 = 1$ , $d_s'neq 1$ (using the cardinality hypothesis on $D$ - and $d_1' neq 1$, but again this follows from the reduced form).



          So with these hypotheses $[x,y] = c_1d_1...c_rd_1'....c_s'd_s'c_r^{-1}....d_1^{-1}c_1^{-1}d_s'^{-1}c_s'^{-1}...d_1'^{-1}$ which is written in reduced form, and is thus $neq 1$. Therefore $[x,y]in Hcap Ksetminus{1}$.



          Apply this to your supposed $Atimes {1}, {1}times B$ to get a contradiction.



          I don't know if there's an easier argument, or a not-too-complicated argument that encapsulates the low cardinality cases, but I guess for these you have to go "by hand" somehow; or perhaps you can adapt this argument to these cases by working a bit more. In any case I didn't want to bother with these cases, and this argument works in most cases and is pretty painless so in any case it's interesting to share






          share|cite|improve this answer











          $endgroup$
















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            3








            3





            $begingroup$

            Let me just replicate YCor's comment in the link that was given in the comments, with more details :




            in a free product $Cast D$ with $C,D$ nontrivial, the intersection of any two nontrivial normal subgroups is nontrivial.




            Indeed, let $H,K$ be two nontrivial normal subgroups of $C*D$. I'll assume for simplicity that $|C|,|D|$ are large enough so that for each $x,y$ there is a nontrivial $z$ with $z^{-1}neq x, zneq y$. For instance $|C|, |D|geq 4$ is good enough (take $x,y$, then there are at most two nontrivial $z$ such that $z=x$ or $z^{-1}=y$ : $x$ and $y^{-1}$; so if $|G|geq 4$, any nontrivial element different from $x$ and $y^{-1}$ works)



            Let me say that $c_1d_1....c_nd_n$ is the reduced form of an element of $C*D$ if $c_iin C, d_jin D$, and the only $c_i,d_j$ allowed to be $1$ are $c_1$ and $d_n$. Clearly, if $x=c_1d_1....c_nd_n$ is the reduced form of $x$, and $ngeq 2$, then $xneq 1$ in $C*D$ ("clearly" here is to be understood as : it's a classical property of free products), moreover, if $n=1$ this is $1$ if and only if $c_1=d_1=1$.



            Now let $x= c_1d_1...c_rd_r in H, y=c'_1d'_1...c'_sd'_s in K$ be nontrivial elements, with obvious notations, written in reduced form. The point will be that $[x,y]in Hcap K$ (this is obvious), and that, up to changing $x,y$ a bit, this can't be the trivial element.



            Now up to conjugation by some element of $C$, one may assume $d_r = 1$ and $c_1neq 1$ (this can be done by the hypothesis on $|C|$ - and $c_r neq 1$, but that follows from the reduced form); and up to conjugation by some element of $D$, $c'_1 = 1$ , $d_s'neq 1$ (using the cardinality hypothesis on $D$ - and $d_1' neq 1$, but again this follows from the reduced form).



            So with these hypotheses $[x,y] = c_1d_1...c_rd_1'....c_s'd_s'c_r^{-1}....d_1^{-1}c_1^{-1}d_s'^{-1}c_s'^{-1}...d_1'^{-1}$ which is written in reduced form, and is thus $neq 1$. Therefore $[x,y]in Hcap Ksetminus{1}$.



            Apply this to your supposed $Atimes {1}, {1}times B$ to get a contradiction.



            I don't know if there's an easier argument, or a not-too-complicated argument that encapsulates the low cardinality cases, but I guess for these you have to go "by hand" somehow; or perhaps you can adapt this argument to these cases by working a bit more. In any case I didn't want to bother with these cases, and this argument works in most cases and is pretty painless so in any case it's interesting to share






            share|cite|improve this answer











            $endgroup$



            Let me just replicate YCor's comment in the link that was given in the comments, with more details :




            in a free product $Cast D$ with $C,D$ nontrivial, the intersection of any two nontrivial normal subgroups is nontrivial.




            Indeed, let $H,K$ be two nontrivial normal subgroups of $C*D$. I'll assume for simplicity that $|C|,|D|$ are large enough so that for each $x,y$ there is a nontrivial $z$ with $z^{-1}neq x, zneq y$. For instance $|C|, |D|geq 4$ is good enough (take $x,y$, then there are at most two nontrivial $z$ such that $z=x$ or $z^{-1}=y$ : $x$ and $y^{-1}$; so if $|G|geq 4$, any nontrivial element different from $x$ and $y^{-1}$ works)



            Let me say that $c_1d_1....c_nd_n$ is the reduced form of an element of $C*D$ if $c_iin C, d_jin D$, and the only $c_i,d_j$ allowed to be $1$ are $c_1$ and $d_n$. Clearly, if $x=c_1d_1....c_nd_n$ is the reduced form of $x$, and $ngeq 2$, then $xneq 1$ in $C*D$ ("clearly" here is to be understood as : it's a classical property of free products), moreover, if $n=1$ this is $1$ if and only if $c_1=d_1=1$.



            Now let $x= c_1d_1...c_rd_r in H, y=c'_1d'_1...c'_sd'_s in K$ be nontrivial elements, with obvious notations, written in reduced form. The point will be that $[x,y]in Hcap K$ (this is obvious), and that, up to changing $x,y$ a bit, this can't be the trivial element.



            Now up to conjugation by some element of $C$, one may assume $d_r = 1$ and $c_1neq 1$ (this can be done by the hypothesis on $|C|$ - and $c_r neq 1$, but that follows from the reduced form); and up to conjugation by some element of $D$, $c'_1 = 1$ , $d_s'neq 1$ (using the cardinality hypothesis on $D$ - and $d_1' neq 1$, but again this follows from the reduced form).



            So with these hypotheses $[x,y] = c_1d_1...c_rd_1'....c_s'd_s'c_r^{-1}....d_1^{-1}c_1^{-1}d_s'^{-1}c_s'^{-1}...d_1'^{-1}$ which is written in reduced form, and is thus $neq 1$. Therefore $[x,y]in Hcap Ksetminus{1}$.



            Apply this to your supposed $Atimes {1}, {1}times B$ to get a contradiction.



            I don't know if there's an easier argument, or a not-too-complicated argument that encapsulates the low cardinality cases, but I guess for these you have to go "by hand" somehow; or perhaps you can adapt this argument to these cases by working a bit more. In any case I didn't want to bother with these cases, and this argument works in most cases and is pretty painless so in any case it's interesting to share







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 hours ago









            YCor

            7,528929




            7,528929










            answered 6 hours ago









            MaxMax

            14.3k11142




            14.3k11142























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                $begingroup$

                Let $f:Atimes Brightarrow C*D$ be an isomorphism where $A,B,C,D$ are non trivial groups. Let $a_0in A$ not trivial, for every $bin B$, $f(a_0)$ commutes with $f(b)$.



                Now we look the proof of the The Corollary 4.1.6 p. 187 of Magnus Karrass and Solitar.



                The first part of this proof asserts that if $f(a_0)$ is contained in the conjugated of a free factor, that is $f(a_0)$ is in $gCg^{-1}$ or is in $gDg^{-1}$ then so is $f(b)$ for every $bin B$. Without restricting the generality, we suppose that $f(a_0)$ and so $f(B)$ are contained in $gCg^{-1}$. Let $b_0$ a non trivial element of $B$ since $B$ is contained in the free conjugated factor $gCg^{-1}$ the same argument shows that $f(A)$ is contained in $gCg^{-1}$ this implies that $C*D$ is contained in a free conjugated factor. Contradiction.



                The second part of the proof shows that if $f(a_0)$ is not contained in a free conjugated factor, then there exists $u_c$ such $f(a_0)$ and $f(b)$ are a power of $u_c$.



                You can express $f(a_0)$ uniquely as a reduced sequence (Theorem 4.1.) this implies that there exists a unique element $u$ with minimal length such that $f(a_0)$ is a power of $u$ and if $f(a_0)$ is a power of $v$, $v$ is a power of $u$. This impllies that every element of $f(B)$ are power of $u$ and $f(B)$ and $B$ are cyclic. A similar argument shows that $f(A)$ and $A$ are cyclic, we deduce that $Atimes B$ is commutative. Contradicition since $C*D$ is not commutative.



                Reference.



                Combinatorial Group theory.



                Magnus, Karrass and Solitar.






                share|cite|improve this answer











                $endgroup$


















                  3












                  $begingroup$

                  Let $f:Atimes Brightarrow C*D$ be an isomorphism where $A,B,C,D$ are non trivial groups. Let $a_0in A$ not trivial, for every $bin B$, $f(a_0)$ commutes with $f(b)$.



                  Now we look the proof of the The Corollary 4.1.6 p. 187 of Magnus Karrass and Solitar.



                  The first part of this proof asserts that if $f(a_0)$ is contained in the conjugated of a free factor, that is $f(a_0)$ is in $gCg^{-1}$ or is in $gDg^{-1}$ then so is $f(b)$ for every $bin B$. Without restricting the generality, we suppose that $f(a_0)$ and so $f(B)$ are contained in $gCg^{-1}$. Let $b_0$ a non trivial element of $B$ since $B$ is contained in the free conjugated factor $gCg^{-1}$ the same argument shows that $f(A)$ is contained in $gCg^{-1}$ this implies that $C*D$ is contained in a free conjugated factor. Contradiction.



                  The second part of the proof shows that if $f(a_0)$ is not contained in a free conjugated factor, then there exists $u_c$ such $f(a_0)$ and $f(b)$ are a power of $u_c$.



                  You can express $f(a_0)$ uniquely as a reduced sequence (Theorem 4.1.) this implies that there exists a unique element $u$ with minimal length such that $f(a_0)$ is a power of $u$ and if $f(a_0)$ is a power of $v$, $v$ is a power of $u$. This impllies that every element of $f(B)$ are power of $u$ and $f(B)$ and $B$ are cyclic. A similar argument shows that $f(A)$ and $A$ are cyclic, we deduce that $Atimes B$ is commutative. Contradicition since $C*D$ is not commutative.



                  Reference.



                  Combinatorial Group theory.



                  Magnus, Karrass and Solitar.






                  share|cite|improve this answer











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Let $f:Atimes Brightarrow C*D$ be an isomorphism where $A,B,C,D$ are non trivial groups. Let $a_0in A$ not trivial, for every $bin B$, $f(a_0)$ commutes with $f(b)$.



                    Now we look the proof of the The Corollary 4.1.6 p. 187 of Magnus Karrass and Solitar.



                    The first part of this proof asserts that if $f(a_0)$ is contained in the conjugated of a free factor, that is $f(a_0)$ is in $gCg^{-1}$ or is in $gDg^{-1}$ then so is $f(b)$ for every $bin B$. Without restricting the generality, we suppose that $f(a_0)$ and so $f(B)$ are contained in $gCg^{-1}$. Let $b_0$ a non trivial element of $B$ since $B$ is contained in the free conjugated factor $gCg^{-1}$ the same argument shows that $f(A)$ is contained in $gCg^{-1}$ this implies that $C*D$ is contained in a free conjugated factor. Contradiction.



                    The second part of the proof shows that if $f(a_0)$ is not contained in a free conjugated factor, then there exists $u_c$ such $f(a_0)$ and $f(b)$ are a power of $u_c$.



                    You can express $f(a_0)$ uniquely as a reduced sequence (Theorem 4.1.) this implies that there exists a unique element $u$ with minimal length such that $f(a_0)$ is a power of $u$ and if $f(a_0)$ is a power of $v$, $v$ is a power of $u$. This impllies that every element of $f(B)$ are power of $u$ and $f(B)$ and $B$ are cyclic. A similar argument shows that $f(A)$ and $A$ are cyclic, we deduce that $Atimes B$ is commutative. Contradicition since $C*D$ is not commutative.



                    Reference.



                    Combinatorial Group theory.



                    Magnus, Karrass and Solitar.






                    share|cite|improve this answer











                    $endgroup$



                    Let $f:Atimes Brightarrow C*D$ be an isomorphism where $A,B,C,D$ are non trivial groups. Let $a_0in A$ not trivial, for every $bin B$, $f(a_0)$ commutes with $f(b)$.



                    Now we look the proof of the The Corollary 4.1.6 p. 187 of Magnus Karrass and Solitar.



                    The first part of this proof asserts that if $f(a_0)$ is contained in the conjugated of a free factor, that is $f(a_0)$ is in $gCg^{-1}$ or is in $gDg^{-1}$ then so is $f(b)$ for every $bin B$. Without restricting the generality, we suppose that $f(a_0)$ and so $f(B)$ are contained in $gCg^{-1}$. Let $b_0$ a non trivial element of $B$ since $B$ is contained in the free conjugated factor $gCg^{-1}$ the same argument shows that $f(A)$ is contained in $gCg^{-1}$ this implies that $C*D$ is contained in a free conjugated factor. Contradiction.



                    The second part of the proof shows that if $f(a_0)$ is not contained in a free conjugated factor, then there exists $u_c$ such $f(a_0)$ and $f(b)$ are a power of $u_c$.



                    You can express $f(a_0)$ uniquely as a reduced sequence (Theorem 4.1.) this implies that there exists a unique element $u$ with minimal length such that $f(a_0)$ is a power of $u$ and if $f(a_0)$ is a power of $v$, $v$ is a power of $u$. This impllies that every element of $f(B)$ are power of $u$ and $f(B)$ and $B$ are cyclic. A similar argument shows that $f(A)$ and $A$ are cyclic, we deduce that $Atimes B$ is commutative. Contradicition since $C*D$ is not commutative.



                    Reference.



                    Combinatorial Group theory.



                    Magnus, Karrass and Solitar.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 5 hours ago

























                    answered 5 hours ago









                    Tsemo AristideTsemo Aristide

                    58k11445




                    58k11445























                        0












                        $begingroup$

                        To be a free product $Cast D$ implies the existence of an action on oriented tree such that $C$ is the stabilizer of some vertex $v_0$ and such that edge stabilizers are trivial (so that nontrivial element fix at most one vertex).



                        Let $(a,b)$ be a nontrivial element of $C$; we can suppose that $aneq 1$ up to switch $A$ and $B$. Then in the Bass-Serre tree, $(a,b)$ fixes a unique vertex $v_0$, and hence this vertex is also fixed by the centralizer of $(a,b)$, and hence $(a,1)$ fixes $v_0$. Applying this to $(a,1)$ shows that $B$ fixes the vertex $v_0$. Choose $1neq b'in B$. It fixes the unique vertex $v_0$ and hence its centralizer fixes $v_0$, so $A$ fixes $v_0$. Finally $G=Atimes B$ fixes $v_0$, so $D=1$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          To be a free product $Cast D$ implies the existence of an action on oriented tree such that $C$ is the stabilizer of some vertex $v_0$ and such that edge stabilizers are trivial (so that nontrivial element fix at most one vertex).



                          Let $(a,b)$ be a nontrivial element of $C$; we can suppose that $aneq 1$ up to switch $A$ and $B$. Then in the Bass-Serre tree, $(a,b)$ fixes a unique vertex $v_0$, and hence this vertex is also fixed by the centralizer of $(a,b)$, and hence $(a,1)$ fixes $v_0$. Applying this to $(a,1)$ shows that $B$ fixes the vertex $v_0$. Choose $1neq b'in B$. It fixes the unique vertex $v_0$ and hence its centralizer fixes $v_0$, so $A$ fixes $v_0$. Finally $G=Atimes B$ fixes $v_0$, so $D=1$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            To be a free product $Cast D$ implies the existence of an action on oriented tree such that $C$ is the stabilizer of some vertex $v_0$ and such that edge stabilizers are trivial (so that nontrivial element fix at most one vertex).



                            Let $(a,b)$ be a nontrivial element of $C$; we can suppose that $aneq 1$ up to switch $A$ and $B$. Then in the Bass-Serre tree, $(a,b)$ fixes a unique vertex $v_0$, and hence this vertex is also fixed by the centralizer of $(a,b)$, and hence $(a,1)$ fixes $v_0$. Applying this to $(a,1)$ shows that $B$ fixes the vertex $v_0$. Choose $1neq b'in B$. It fixes the unique vertex $v_0$ and hence its centralizer fixes $v_0$, so $A$ fixes $v_0$. Finally $G=Atimes B$ fixes $v_0$, so $D=1$.






                            share|cite|improve this answer









                            $endgroup$



                            To be a free product $Cast D$ implies the existence of an action on oriented tree such that $C$ is the stabilizer of some vertex $v_0$ and such that edge stabilizers are trivial (so that nontrivial element fix at most one vertex).



                            Let $(a,b)$ be a nontrivial element of $C$; we can suppose that $aneq 1$ up to switch $A$ and $B$. Then in the Bass-Serre tree, $(a,b)$ fixes a unique vertex $v_0$, and hence this vertex is also fixed by the centralizer of $(a,b)$, and hence $(a,1)$ fixes $v_0$. Applying this to $(a,1)$ shows that $B$ fixes the vertex $v_0$. Choose $1neq b'in B$. It fixes the unique vertex $v_0$ and hence its centralizer fixes $v_0$, so $A$ fixes $v_0$. Finally $G=Atimes B$ fixes $v_0$, so $D=1$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            YCorYCor

                            7,528929




                            7,528929






























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