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I want presentation of non-Abelian group $(C_3\times C_3)\rtimes C_3$ of order $27.$ Actually it is $$(C_3\times C_3)\rtimes C_3\cong \{\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}\;|\; a,b,c\in \Bbb{Z}_3\}$$. But i want its presentation.

There is link Classify groups of order 27 which classifies all group of order $27.$ In this link presentation is given as $\langle a,b\mid a^3,b^3,[a,b]=[a,b]^a=[a,b]^b\rangle$ but i did't get this. Please help me so that i can write all $27$ elements of group. I also tried in GAP as below

gap> p:=3;

3

gap> N:=CyclicGroup(p);

gap> K:=DirectProduct(N,N);

gap> H:=CyclicGroup(p);

gap> hom:=AllHomomorphisms(H,AutomorphismGroup(K));

[ [ f1 ] -> [ IdentityMapping( ) ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1*f2^2, f2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1, f1^2*f2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1*f2, f2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1^2*f2, f1^2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1, f1*f2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f2^2, f1*f2^2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f2, f1^2*f2^2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1^2*f2^2, f1 ] ] ]

gap> hom:=AllHomomorphisms(H,AutomorphismGroup(K)); [ [ f1 ] -> [ IdentityMapping( ) ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1*f2^2, f2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1, f1^2*f2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1*f2, f2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1^2*f2, f1^2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1, f1*f2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f2^2, f1*f2^2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f2, f1^2*f2^2 ] ], [ f1 ] -> [ Pcgs([ f1, f2 ]) -> [ f1^2*f2^2, f1 ] ] ]

gap> g:=SemidirectProduct(H,hom[1],K);

gap> g:=SemidirectProduct(H,hom[1],K);

gap> StructureDescription(g);

"C3 x C3 x C3"

gap> g:=SemidirectProduct(H,hom[2],K);

gap> StructureDescription(g);

"(C3 x C3) : C3"

gap> IsAbelian(g);

false gap> Size(g);

27 gap> Elements(g);

[ of ..., f1, f2, f3, f1^2, f1*f2, f1*f3, f2^2, f2*f3, f3^2, f1^2*f2, f1^2*f3, f1*f2^2, f1*f2*f3, f1*f3^2, f2^2*f3, f2*f3^2, f1^2*f2^2, f1^2*f2*f3, f1^2*f3^2, f1*f2^2*f3, f1*f2*f3^2, f2^2*f3^2, f1^2*f2^2*f3, f1^2*f2*f3^2, f1*f2^2*f3^2, f1^2*f2^2*f3^2 ]

gap> t:=Elements(g); [ of ..., f1, f2, f3, f1^2, f1*f2, f1*f3, f2^2, f2*f3, f3^2, f1^2*f2, f1^2*f3, f1*f2^2, f1*f2*f3, f1*f3^2, f2^2*f3, f2*f3^2, f1^2*f2^2, f1^2*f2*f3, f1^2*f3^2, f1*f2^2*f3, f1*f2*f3^2, f2^2*f3^2, f1^2*f2^2*f3, f1^2*f2*f3^2, f1*f2^2*f3^2, f1^2*f2^2*f3^2 ]

So please help me to find a representation so that i can find all elements easily thanks.

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  • $\begingroup$ Are you sure that the semidirect product is isomorphic to the matrix group (Heisenberg group) as you write it? $\endgroup$ – user8268 Apr 12 '18 at 5:56
  • $\begingroup$ Yes it's.................... $\endgroup$ – neelkanth Apr 12 '18 at 5:58
  • $\begingroup$ Hmm, but the answer you linked seems to disagree - it lists Heisenberg group and the semidirect product as two different groups $\endgroup$ – user8268 Apr 12 '18 at 6:12
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Write $a$ and $b$ as generators of the elementary Abelian subgroup $A$ of order $9$. They satisfy $a^3=b^3=1$ and $ab=ba$. Let $c$ be a further generator. It has order $3$. It normalises $A$, and acts as an automorphism of order $3$ on $A$. One such maps $a$ to $a$ and $b$ to $ab$. So add relations $c^3=1$, $ca=ac$ and $cb=abc$. So overall we have $$a^3=b^3=c^3=1,\qquad ba=ab,\qquad ca=ac,\qquad cb=abc.$$

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  • $\begingroup$ 's Can we find it's two element presentation as mention in question $\endgroup$ – neelkanth Apr 12 '18 at 9:51
  • $\begingroup$ How to verify it so that I can use this presentation .... $\endgroup$ – neelkanth Apr 12 '18 at 11:47
  • $\begingroup$ $cb = abc$ is equivalent to $cbc^{-1}b^{-1} = a$, so it's essentialy same as presentation with two generators except that we add excessive generator $a = [c, b]$. $\endgroup$ – xsnl Apr 12 '18 at 14:59

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