IB DP Further Mathematics 4.11 Subgroups, proper subgroups HL Paper 2

Question

The relation \({R_1}\) is defined for \(a,b \in {\mathbb{Z}^ + }\) by \(a{R_1}b\) if and only if \(n\left| {({a^2} – {b^2})} \right.\) where \(n\) is a fixed positive integer.

  (i)     Show that \({R_1}\) is an equivalence relation.

  (ii)     Determine the equivalence classes when \(n = 8\) .

[11]
A.a.

Consider the group \(\left\{ {G, * } \right\}\) and let \(H\) be a subset of \(G\) defined by

\(H = \left\{ {x \in G} \right.\) such that \(x * a = a * x\) for all \(a \in \left. G \right\}\) .

Show that \(\left\{ {H, * } \right\}\) is a subgroup of \(\left\{ {G, * } \right\}\) .

[12]
B.

The relation \({R_2}\) is defined for \(a,b \in {\mathbb{Z}^ + }\) by \(a{R_2}b\) if and only if \((4 + \left| {a – b} \right|)\) is the square of a positive integer. Show that \({R_2}\) is not transitive.

[3]
B.b.
Answer/Explanation

Markscheme

(i)     Since \({a^2} – {a^2} = 0\) is divisible by n, it follows that \(a{R_1}a\) so \({R_1}\) is reflexive.     A1

\(a{R_1}b \Rightarrow {a^2} – {b^2}\) divisible by \(n \Rightarrow {b^2} – {a^2}\) divisible by \(n \Rightarrow b{R_1}a\) so

symmetric.     A1

\(a{R_1}b\) and \(b{R_1}c \Rightarrow {a^2} – {b^2} = pn\) and \({b^2} – {c^2} = qn\)    A1

\(({a^2} – {b^2}) + ({b^2} – {c^2}) = pn + qn\)     M1

so \({a^2} – {c^2} = (p + q)n \Rightarrow a{R_1}c\)     A1

Therefore \({R_1}\) is transitive.

It follows that \({R_1}\) is an equivalence relation.     AG

(ii)     When \(n = 8\) , the equivalence classes are

\(\left\{ {1,3,5,7,9, \ldots } \right\}\) , i.e. the odd integers     A2

\(\left\{ {2,6,10,14, \ldots } \right\}\)     A2

and \(\left\{ {4,8,12,16, \ldots } \right\}\)     A2

Note: If finite sets are shown award A1A1A1.

[11 marks]

A.a.

Associativity follows since G is associative.     A1

Closure: Let \(x,y \in H\) so \(ax = xa\) , \(ay = ya\) for \(a \in G\)     M1

Consider \(axy = xay = xya \Rightarrow xy \in H\)     M1A1

The identity \(e \in H\) since \(ae = ea\) for \(a \in G\)     A2

Inverse: Let \(x \in H\) so \(ax = xa\) for \(a \in G\)

Then

\({x^{ – 1}}a = {x^{ – 1}}ax{x^{ – 1}}\)     M1A1

\( = {x^{ – 1}}xa{x^{ – 1}}\)     M1

\( = a{x^{ – 1}}\)     A1

so \( \Rightarrow {x^{ – 1}} \in H\)     A1

The four group axioms are satisfied so \(H\) is a subgroup.     R1

[12 marks]

B.

Attempt to find a counter example.     (M1)

We note that \(1{R_2}6\) and \(6{R_2}11\) but 1 not \({R_2}11\) .     A2

Note: Accept any valid counter example.

The relation is not transitive.     AG

[3 marks]

B.b.

Question

(i)     Draw the Cayley table for the set \(S = \left\{ {0,1,2,3,4,\left. 5 \right\}} \right.\) under addition modulo six \(({ + _6})\) and hence show that \(\left\{ {S, + \left. {_6} \right\}} \right.\) is a group.

(ii)     Show that the group is cyclic and write down its generators.

(iii)     Find the subgroup of \(\left\{ {S, + \left. {_6} \right\}} \right.\) that contains exactly three elements.

[11]
a.

Prove that a cyclic group with exactly one generator cannot have more than two elements.

[4]
b.

\(H\) is a group and the function \(\Phi :H \to H\) is defined by \(\Phi (a) = {a^{ – 1}}\) , where \({a^{ – 1}}\) is the inverse of a under the group operation. Show that \(\Phi \) is an isomorphism if and only if H is Abelian.

[9]
c.
Answer/Explanation

Markscheme

(i)

the table is closed    A1

the identity is \(0\)     A1

\(0\) is in every row and column once so each element has a unique inverse     A1

addition is associative     A1

therefore \(\left\{ {S, + \left. {_6} \right\}} \right.\) is a group     R1

(ii)     \(1 + 1 + 1 + 1 + 1 + 1 = 0\)     M1

\(1 + 1 + 1 + 1 + 1 = 5\)

\(1 + 1 + 1 + 1 = 4\)

\(1 + 1 + 1 = 3\)

\(1 + 1 = 2\)

so \(1\) is a generator of \(\left\{ {S, + \left. {_6} \right\}} \right.\) and the group is cyclic     A1

(since \(5\) is the additive inverse of \(1\)) \(5\) is also a generator     A1 

(iii)     \(\left\{ {0,2,\left. 4 \right\}} \right.\)     A1 

[11 marks]

a.

if \(a\) is a generator of group \((G, * )\) then so is \({a^{ – 1}}\)     A1

if \((G, * )\) has exactly one generator \(a\) then \(a = {a^{ – 1}}\)     A1

so \({a^2} = e\) and \(G = \left\{ {e,\left. a \right\}} \right.\) \(\left\{ {\left. e \right\}} \right.\)     A1R1

so cyclic group with exactly one generator cannot have more than two elements     AG

[4 marks]

b.

every element of a group has a unique inverse so \(\Phi \) is a bijection     A1

\(\Phi (ab) = {(ab)^{ – 1}} = {b^{ – 1}}{a^{ – 1}}\)     M1A1

if \(H\) is Abelian then it follows that

\({b^{ – 1}}{a^{ – 1}} = {a^{ – 1}}{b^{ – 1}} = \Phi (a)\Phi (b)\)     A1

so \(\Phi \) is an isomorphism     R1

if \(\Phi \) is an isomorphism, then     M1

for all \(a,b \in H\) , \(\Phi (ab) = \Phi (a)\Phi (b)\)     M1

\({(ab)^{ – 1}} = {a^{ – 1}}{b^{ – 1}}\)

\( \Rightarrow {b^{ – 1}}{a^{ – 1}} = {a^{ – 1}}{b^{ – 1}}\)    A1

so \(H\) is Abelian     R1

[9 marks]

c.

Question

The function \(f:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) is defined by \(\boldsymbol{X} \mapsto \boldsymbol{AX}\) , where \(\boldsymbol{X} = \left[ \begin{array}{l}
x\\
y
\end{array} \right]\) and \(\boldsymbol{A} = \left[ \begin{array}{l}
a\\
c
\end{array} \right.\left. \begin{array}{l}
b\\
d
\end{array} \right]\) where \(a\) , \(b\) , \(c\) , \(d\) are all non-zero.

Consider the group \(\left\{ {S,{ + _m}} \right\}\) where \(S = \left\{ {0,1,2 \ldots m – 1} \right\}\) , \(m \in \mathbb{N}\) , \(m \ge 3\) and \({ + _m}\) denotes addition modulo \(m\) .

Show that \(f\) is a bijection if \(\boldsymbol{A}\) is non-singular.

[7]
A.a.

Suppose now that \(\boldsymbol{A}\) is singular.

  (i)     Write down the relationship between \(a\) , \(b\) , \(c\) , \(d\) .

  (ii)     Deduce that the second row of \(\boldsymbol{A}\) is a multiple of the first row of \(\boldsymbol{A}\) .

  (iii)     Hence show that \(f\) is not a bijection.

[5]
A.b.

Show that \(\left\{ {S,{ + _m}} \right\}\) is cyclic for all m .

[3]
B.a.

Given that \(m\) is prime,

  (i)     explain why all elements except the identity are generators of \(\left\{ {S,{ + _m}} \right\}\) ;

  (ii)     find the inverse of \(x\) , where x is any element of \(\left\{ {S,{ + _m}} \right\}\) apart from the identity;

  (iii)     determine the number of sets of two distinct elements where each element is the inverse of the other.

[7]
B.b.

Suppose now that \(m = ab\) where \(a\) , \(b\) are unequal prime numbers. Show that \(\left\{ {S,{ + _m}} \right\}\) has two proper subgroups and identify them.

[3]
B.c.
Answer/Explanation

Markscheme

recognizing that the function needs to be injective and surjective     R1

Note: Award R1 if this is seen anywhere in the solution.

injective:

let \(\boldsymbol{U}, \boldsymbol{V} \in ^\circ  \times ^\circ \) be 2-D column vectors such that \(\boldsymbol{AU} = \boldsymbol{AV}\)     M1

\({\boldsymbol{A}^{ – 1}}\boldsymbol{AU} = {\boldsymbol{A}^{ – 1}}\boldsymbol{AV}\)     M1

\(\boldsymbol{U} = \boldsymbol{V}\)     A1

this shows that \(f\) is injective

surjective:

let \(W \in ^\circ  \times ^\circ \)     M1

then there exists \(\boldsymbol{Z} = {\boldsymbol{A}^{ – 1}}\boldsymbol{W} \in ^\circ  \times ^\circ \) such that \(\boldsymbol{AZ} = \boldsymbol{W}\)     M1A1

this shows that \(f\) is surjective

therefore \(f\) is a bijection     AG

[7 marks]

A.a.

(i)     the relationship is \(ad = bc\)     A1

(ii)     it follows that \(\frac{c}{a} = \frac{d}{b} = \lambda \) so that \((c,d) = \lambda (a,b)\)     A1 

(iii)     EITHER

let \(\boldsymbol{W} = \left[ \begin{array}{l}
p\\
q
\end{array} \right]\) be a 2-D vector

then \(\boldsymbol{AW} = \left[ \begin{array}{l}
a\\
\lambda a
\end{array} \right.\left. \begin{array}{l}
b\\
\lambda b
\end{array} \right]\left[ \begin{array}{l}
p\\
q
\end{array} \right]\)     M1

\( = \left[ \begin{array}{l}
ap + bq\\
\lambda (ap + bq)
\end{array} \right]\)    A1

the image always satisfies \(y = \lambda x\) so \(f\) is not surjective and therefore not a bijection     R1

OR

consider

\(\left[ {\begin{array}{*{20}{c}}
  a&b \\
  {\lambda a}&{\lambda b}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  b \\
  0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {ab} \\
  {\lambda ab}
\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}
  a&b \\
  {\lambda a}&{\lambda b}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  0 \\
  a
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {ab} \\
  {\lambda ab}
\end{array}} \right]\)

this shows that \(f\) is not injective and therefore not a bijection     R1 

[5 marks]

A.b.

the identity element is \(0\)     R1

consider, for \(1 \le r \le m\) ,

using \(1\) as a generator     M1

\(1\) combined with itself \(r\) times gives \(r\) and as \(r\) increases from \(1\) to m, the group is generated ending with \(0\) when \(r = m\)     A1

it is therefore cyclic     AG

[3 marks]

B.a.

(i)     by Lagrange the order of each element must be a factor of \(m\) and if \(m\) is prime, its only factors are \(1\) and \(m\)     R1

since 0 is the only element of order \(1\), all other elements are of order \(m\) and are therefore generators     R1

(ii)     since \(x{ + _m}(m – x) = 0\)     (M1)

the inverse of x is \((m – x)\)     A1

(iii)     consider


     M1A1

there are \(\frac{1}{2}(m – 1)\) inverse pairs     A1 N1

Note: Award M1 for an attempt to list the inverse pairs, A1 for completing it correctly and A1 for the final answer.

[7 marks]

B.b.

since \(a\), \(b\) are unequal primes the only factors of \(m\) are \(a\) and \(b\)

there are therefore only subgroups of order \(a\) and \(b\)     R1

they are

\(\left\{ {0,a,2a, \ldots ,(b – 1)a} \right\}\)     A1

\(\left\{ {0,b,2b, \ldots ,(a – 1)b} \right\}\)     A1

[3 marks]

B.c.

Question

Let \(f\) be a homomorphism of a group \(G\) onto a group \(H\).

Show that if \(e\) is the identity in \(G\), then \(f(e)\) is the identity in \(H\).

[2]
a.

Show that if \(x\) is an element of \(G\), then \(f({x^{ – 1}}) = {\left( {f(x)} \right)^{ – 1}}\).

[2]
b.

Show that if \(G\) is Abelian, then \(H\) must also be Abelian.

[5]
c.

Show that if \(S\) is a subgroup of \(G\), then \(f(S)\) is a subgroup of \(H\).

[4]
d.
Answer/Explanation

Markscheme

\(f(a) = f(ae) = f(a)f(e)\)     M1A1

hence \(f(e)\) is the identity in \(H\)     AG

a.

\(e’ = f(e)\)

\( = f(x{x^{ – 1}})\)     M1

\( = f(x)f({x^{ – 1}})\)     A1

hence \(f({x^{ – 1}}) = {\left( {f(x)} \right)^{ – 1}}\)     AG

b.

let \(a’,{\text{ }}b’ \in H\), we need to show that \(a’b’ = b’a’\)     (M1)

since \(f\) is onto \(H\) there exists \(a,{\text{ }}b \in G\) such that \(f(a) = a’\)

and \(f(b) = b’\)     (M1)

now \(a’b’ = f(a)f(b) = f(ab)\)     A1

since \(f(ab) = f(ba)\)     M1

\(f(ba) = f(b)f(a) = b’a’\)     A1

hence Abelian     AG

c.

METHOD 1

\(e’ = f(e)\) and \(f({x^{ – 1}}) = {\left( {f(x)} \right)^{ – 1}}\) from above     A1A1

let \(f(a)\) and \(f(b)\) be two elements in \(f(S)\)

then \(f(a)f(b) = f(ab)\)     M1

\( \Rightarrow f(a)f(b) \in f(S)\)     A1

hence closed under the operation of \(H\)

\(f(S)\) is a subgroup of \(H\)     AG

METHOD 2

\(f(S)\) contains the identity, so is non empty     A1

Suppose \(f(a),{\text{ }}f(b) \in f(S)\)

Consider \(f(a)f{(b)^{ – 1}}\)     M1

\( = f(a)f({b^{ – 1}})\)     (from (b))     A1

\( = f(a{b^{ – 1}})\)     (homomorphism)     A1

\( \in f(S)\) as \(a{b^{ – 1}} \in H\)

So \(f(S)\) is a subgroup of \(H\) (by a subgroup theorem)     AG

d.

Question

The binary operation multiplication modulo \(9\), denoted by \({ \times _9}\) , is defined on the set \(S = \left\{ {1,2,3,4,5,6,7,8} \right\}\) .

Copy and complete the following Cayley table.



[3]
a.

Show that \(\left\{ {S,{ \times _9}} \right\}\) is not a group.

[1]
b.

Prove that a group \(\left\{ {G,{ \times _9}} \right\}\) can be formed by removing two elements from the set \(S\) .

[5]
c.

(i)     Find the order of all the elements of \(G\) .

(ii)     Write down all the proper subgroups of \(\left\{ {G,{ \times _9}} \right\}\) .

(iii)     Determine the coset containing the element \(5\) for each of the subgroups in part (ii).

[8]
d.

Solve the equation \(4{ \times _9}x{ \times _9}x = 1\) .

[3]
e.
Answer/Explanation

Markscheme

     A3

Note: Award A2 if one error, A1 if two errors and A0 if three or more errors.

[3 marks]

a.

any valid reason,     R1

e.g.  not closed

\(3\) or \(6\) has no inverse,

it is not a Latin square

[1 mark]

b.

remove \(3\) and \(6\)     A1

for the remaining elements,

the table is closed     R1

associative because multiplication is associative     R1

the identity is \(1\)     A1

every element has an inverse, (\(2\), \(5\)) and (\(4\), \(7\)) are inverse pairs and \(8\) (and \(1\)) are self-inverse     A1

thus it is a group     AG

[5 marks]

c.

(i)     the orders are

     A3

Note: Award A2 if one error, A1 if two errors and A0 if three or more errors. 

(ii)     the proper subgroups are

\(\left\{ {1,8} \right\}\)     A1

\(\left\{ {1,4,7} \right\}\)     A1

Note: Do not penalize inclusion of \(\left\{ 1 \right\}\) .

(iii)     the cosets are \(\left\{ {5,4} \right\}\)     (M1)A1

\(\left\{ {5,2,8} \right\}\)     A1

[8 marks]

d.

\(x{ \times _9}x = 7\)     (A1)

\(x = 4,5\)     A1A1

[3 marks]

e.
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