GroupTheory/QuasicyclicGroup - Maple Help

GroupTheory

 QuasicyclicGroup
 construct a quasicyclic group for a given prime

 Calling Sequence QuasicyclicGroup( p )

Parameters

 p - a prime number

Options

 • formopt : option of the form form = "multiplicative" or form = "additive" (the default)

Description

 • A quasicyclic group is an infinite abelian group $G$ in which each element has order a power of a single prime number $p$, and such that every proper subgroup of $G$ is a finite cyclic $p$-group. They are also called Prüfer $p$-groups.
 • Abstractly, a quasicyclic group can be defined as a direct limit of the system of finite cyclic groups of the form ${C}_{{p}^{k}}$, for positive integers $k$, in which ${C}_{{p}^{k}}$ is embedded naturally in ${C}_{{p}^{k+1}}$.
 • The QuasicyclicGroup( p ) calling sequence constructs a quasicyclic p-group, where p is a prime number. By default, an additive representation that models the Sylow p-subgroup of the quotient group $ℚ/ℤ$ is generated. A multiplicative version of the group can be realized by passing the option form = "multiplicative".

 • The default form of a quasicyclic $p$-group $G$ is an additive group that represents the Sylow $p$-subgroup of the quotient group $ℚ/ℤ$ of the additive group of rationals by the integers. Formally, the elements of $G$ are cosets $q+ℤ$ where $q$ is a rational number whose denominator is a power of the prime $p$. Maple, however, uses rational representatives as group elements, so that the group operation is ordinary addition of rationals modulo $1$. That is, two rationals represent the same group element if their difference is an integer. In particular, any integer is a representative of the identity element of the group.
 • Each rational of the form $\frac{k}{{p}^{m}}$, where $k$ and $m$ are integers, has an unique equivalent representative where $k$ and $m$ satisfy $0\le m$ and $k$ is a non-negative integer with $k<{p}^{m}$.
 • The Operations module for the group implements these conventions. Moreover, the CanonicalForm method of the group returns the unique canonical form of any rational representative of a group element.

Multiplicative Quasicyclic Groups

 • An alternative multiplicative form of a quasicyclic $p$-group is the group of complex $p$-power roots of unity. These are the complex numbers of the form ${ⅇ}^{\frac{2Ik\mathrm{Pi}}{{p}^{n}}}$, where $n$ is a non-negative integer, and $k$ is a positive integer such that $k<{p}^{n}$. The group operation is then just ordinary multiplication of complex numbers.
 • Other more complicated expressions can also represent complex roots of unity. You can use the CanonicalForm method of a quasicyclic group to obtain an expression in the form above, subject to simplifications automatically performed by the exp function.
 • The Operations module for the group implements these conventions.

Subgroups of Quasicyclic Groups

 • Since a proper subgroup of a quasicyclic group is not itself quasicyclic (rather, it is a finite cyclic group of prime power order), in order that its elements remain elements of the parent quasicyclic group, it is represented as a QuasicyclicSubgroup object. The group operations are inherited from the parent quasicyclic group. (In fact, full quasicyclic groups are also represented as QuasicyclicSubgroup objects.)

Examples

 > $\mathrm{with}\left(\mathrm{GroupTheory}\right):$
 > $G≔\mathrm{QuasicyclicGroup}\left(5\right)$
 ${G}{≔}{{ℤ}}_{{{5}}^{{\mathrm{\infty }}}}$ (1)

Quasicyclic groups are of type QuasicyclicGroup, and also of type QuasicyclicSubgroup.

 > $\mathrm{type}\left(G,'\mathrm{QuasicyclicGroup}'\right)$
 ${\mathrm{true}}$ (2)
 > $\mathrm{type}\left(G,'\mathrm{QuasicyclicSubgroup}'\right)$
 ${\mathrm{true}}$ (3)
 > $\mathrm{type}\left(G,'\mathrm{Group}'\right)$
 ${\mathrm{true}}$ (4)

Not only are quasicyclic groups not finite:

 > $\mathrm{IsFinite}\left(G\right)$
 ${\mathrm{false}}$ (5)

They are not even finitely generated.

 > $\mathrm{IsFinitelyGenerated}\left(G\right)$
 ${\mathrm{false}}$ (6)
 > $\mathrm{GroupOrder}\left(G\right)$
 ${\mathrm{\infty }}$ (7)

A quasicyclic group is abelian, but not cyclic.

 > $\mathrm{IsAbelian}\left(G\right)$
 ${\mathrm{true}}$ (8)
 > $\mathrm{IsCyclic}\left(G\right)$
 ${\mathrm{false}}$ (9)

Elements of an additive quasicyclic $p$-group are rationals with denominator a power of $p$.

 > $\frac{12}{25}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}∈\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}G$
 ${\mathrm{true}}$ (10)
 > $\mathrm{ElementOrder}\left(\frac{12}{25},G\right)$
 ${25}$ (11)
 > $\frac{3}{5}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}∈\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}G$
 ${\mathrm{true}}$ (12)

The group operation is rational addition modulo $1$.

 > $\frac{12}{25}+\frac{3}{5}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}∈\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}G$
 ${\mathrm{true}}$ (13)
 > $\mathrm{CanonicalForm}\left(\frac{12}{25}+\frac{3}{5},G\right)$
 $\frac{{2}}{{25}}$ (14)
 > $\mathrm{Operations}\left(G\right):-\mathrm{.}\left(\frac{12}{25},\frac{3}{5}\right)$
 $\frac{{2}}{{25}}$ (15)

The rational number $\frac{2}{3}$ is not a member of $G$ because its denominator is not a power of $5$.

 > $\frac{2}{3}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}∈\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}G$
 ${\mathrm{false}}$ (16)

Every integer belongs to $G$ and represents the group identity.

 > $7\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}∈\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}G$
 ${\mathrm{true}}$ (17)
 > $\mathrm{CanonicalForm}\left(7,G\right)$
 ${0}$ (18)

Every element of $G$ has order a power of $5$.

 > $\mathrm{IsPGroup}\left(G\right)$
 ${\mathrm{true}}$ (19)
 > $\mathrm{PGroupPrime}\left(G\right)$
 ${5}$ (20)

Finitely generated subgroups of $G$ are finite and cyclic.

 > $H≔\mathrm{Subgroup}\left(\left[\frac{12}{125},\frac{3}{5}\right],G\right)$
 ${H}{≔}{{ℤ}}_{{{5}}^{{\mathrm{\infty }}}}$ (21)
 > $\mathrm{type}\left(H,'\mathrm{QuasicyclicGroup}'\right)$
 ${\mathrm{false}}$ (22)
 > $\mathrm{type}\left(H,'\mathrm{QuasicyclicSubgroup}'\right)$
 ${\mathrm{true}}$ (23)
 > $\mathrm{IsFinite}\left(H\right)$
 ${\mathrm{true}}$ (24)
 > $\mathrm{IsCyclic}\left(H\right)$
 ${\mathrm{true}}$ (25)
 > $\mathrm{GroupOrder}\left(H\right)$
 ${125}$ (26)

The subgroup lattice of a quasicyclic group is a chain, infinite in length. However, we can visualize the subgroup lattice of finite subgroups of quasicyclic groups.

 > $\mathrm{DrawSubgroupLattice}\left(H\right)$

An additive quasicyclic $p$-group is isomorphic to the multiplicative quasicyclic $p$-group (for the same prime $p$).

 > $M≔\mathrm{QuasicyclicGroup}\left(5,'\mathrm{form}'="multiplicative"\right)$
 ${M}{≔}{{C}}_{{{5}}^{{\mathrm{\infty }}}}$ (27)

The assign option to the AreIsomorphic command affords you the ability to obtain an explicit isomorphism.

 > $\mathrm{AreIsomorphic}\left(G,M,'\mathrm{assign}'='\mathrm{iso}'\right)$
 ${\mathrm{true}}$ (28)
 > $\mathrm{iso}\left(\frac{3}{25}\right)$
 ${{ⅇ}}^{\frac{{6}{}{I}}{{25}}{}{\mathrm{\pi }}}$ (29)
 > $H≔\mathrm{Subgroup}\left(\left[\mathrm{iso}\left(\frac{4}{25}\right)\right],M\right)$
 ${H}{≔}{{C}}_{{25}}$ (30)
 > $\mathrm{IsSubgroup}\left(H,M\right)$
 ${\mathrm{true}}$ (31)
 > $\mathrm{IsSubgroup}\left(H,G\right)$
 ${\mathrm{false}}$ (32)
 > $\mathrm{GroupOrder}\left(H\right)$
 ${25}$ (33)
 > $\mathrm{IdentifySmallGroup}\left(H\right)$
 ${25}{,}{1}$ (34)

Check that H is isomorphic to the cyclic permutation group of the same order.

 > $\mathrm{AreIsomorphic}\left(H,\mathrm{CyclicGroup}\left(25\right)\right)$
 ${\mathrm{true}}$ (35)
 > $L≔\mathrm{Subgroup}\left(\left\{\frac{4}{25}\right\},G\right)$
 ${L}{≔}{{ℤ}}_{{{5}}^{{\mathrm{\infty }}}}$ (36)
 > $\mathrm{AreIsomorphic}\left(H,L\right)$
 ${\mathrm{true}}$ (37)

The elements of $H$ and $L$ are distinct, though they are isomorphic.

 > $\mathrm{Elements}\left(H\right)$
 $\left\{{1}{,}{-}{{ⅇ}}^{\frac{{I}}{{5}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{I}}{{25}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{3}{}{I}}{{5}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{3}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{7}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{9}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{11}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{13}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{17}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{19}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{21}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{-}{{ⅇ}}^{\frac{{23}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{2}{}{I}}{{5}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{2}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{4}{}{I}}{{5}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{4}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{6}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{8}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{12}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{14}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{16}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{18}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{22}{}{I}}{{25}}{}{\mathrm{\pi }}}{,}{{ⅇ}}^{\frac{{24}{}{I}}{{25}}{}{\mathrm{\pi }}}\right\}$ (38)
 > $\mathrm{Elements}\left(L\right)$
 $\left\{{0}{,}\frac{{1}}{{5}}{,}\frac{{1}}{{25}}{,}\frac{{2}}{{5}}{,}\frac{{2}}{{25}}{,}\frac{{3}}{{5}}{,}\frac{{3}}{{25}}{,}\frac{{4}}{{5}}{,}\frac{{4}}{{25}}{,}\frac{{6}}{{25}}{,}\frac{{7}}{{25}}{,}\frac{{8}}{{25}}{,}\frac{{9}}{{25}}{,}\frac{{11}}{{25}}{,}\frac{{12}}{{25}}{,}\frac{{13}}{{25}}{,}\frac{{14}}{{25}}{,}\frac{{16}}{{25}}{,}\frac{{17}}{{25}}{,}\frac{{18}}{{25}}{,}\frac{{19}}{{25}}{,}\frac{{21}}{{25}}{,}\frac{{22}}{{25}}{,}\frac{{23}}{{25}}{,}\frac{{24}}{{25}}\right\}$ (39)

Since quasicyclic groups are infinite, it is not possible to compute all of their elements.

 > $\mathrm{Elements}\left(G\right)$

Similarly, you can iterate over the elements of a quasicyclic group but, as the group is infinite, you need to provide a termination condition, as illustrated in the following example.

 > $\mathbf{for}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}g\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}G\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{do}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{if}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}1000<\mathrm{denom}\left(g\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{then}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{break}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end if}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end do}:$

On the other hand, iterating over the elements of a finite subgroup of a quasicyclic group terminates.

 > $\left\{\mathrm{seq}\right\}\left(\mathrm{ElementOrder}\left(x,L\right),x=L\right)$
 $\left\{{1}{,}{5}{,}{25}\right\}$ (40)

You can convert a finite quasicyclic subgroup to a permutation group, a finitely presented group, or to a Cayley table group.

 > $\mathrm{PermutationGroup}\left(L\right)$
 $⟨\left({1}{,}{3}{,}{5}{,}{7}{,}{9}{,}{2}{,}{10}{,}{11}{,}{12}{,}{13}{,}{4}{,}{14}{,}{15}{,}{16}{,}{17}{,}{6}{,}{18}{,}{19}{,}{20}{,}{21}{,}{8}{,}{22}{,}{23}{,}{24}{,}{25}\right)⟩$ (41)
 > $\mathrm{CayleyTableGroup}\left(L\right)$
 ${\mathrm{< a Cayley table group with 25 elements >}}$ (42)
 > $\mathrm{FPGroup}\left(L\right)$
 $⟨{}{\mathrm{_g}}{}{\mid }{}{{\mathrm{_g}}}^{{25}}{}⟩$ (43)

As quasicyclic groups have no maximal subgroups, they are equal to their Frattini subgroups.

 > $F≔\mathrm{FrattiniSubgroup}\left(M\right)$
 ${F}{≔}{{C}}_{{{5}}^{{\mathrm{\infty }}}}$ (44)
 > $\mathrm{IsSubgroup}\left(M,F\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{and}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{IsSubgroup}\left(F,M\right)$
 ${\mathrm{true}}$ (45)

A multiplicative quasicyclic $2$-group contains the group generated by the imaginary unit.

 > $T≔\mathrm{QuasicyclicGroup}\left(2,'\mathrm{form}'="multiplicative"\right)$
 ${T}{≔}{{C}}_{{{2}}^{{\mathrm{\infty }}}}$ (46)
 > $U≔\mathrm{Subgroup}\left(\left[I\right],T\right)$
 ${U}{≔}{{C}}_{{4}}$ (47)
 > $\mathrm{IsSubgroup}\left(U,T\right)$
 ${\mathrm{true}}$ (48)
 > $\mathrm{GroupOrder}\left(U\right)$
 ${4}$ (49)
 > $\mathrm{Elements}\left(U\right)$
 $\left\{{-1}{,}{1}{,}{-I}{,}{I}\right\}$ (50)

Quasicyclic groups for different primes are, of course, non-isomorphic.

 > $\mathrm{AreIsomorphic}\left(T,M\right)$
 ${\mathrm{false}}$ (51)
 > $\mathrm{AreIsomorphic}\left(G,T\right)$
 ${\mathrm{false}}$ (52)