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Mathematics LibreTexts

Assignment 3

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Exercise 1

For the following, either show that G is a group with a given operation or list the condition(s) that fail.

  1. G=R with the binary operation ab=a+bab for a,bR.
  2.  G=3Z={3m|mZ} with addition.
  3.  G={4,8,12,16} with multiplication mod(20).
Answer

1.  LetG:=R, and let a,b,cG. Let an operation ) onG be defined asab:=a+bab. \\

Is (G,) closed? () is comprised of (+), (), and () over real numbers. These operations are closed over R. Therefore, sinceG=R, (G,) is closed.

Is(G,) associative? Consider the following:a(bc)=a(b+cbc),=a+(b+cbc)a(b+cbc),=a+b+cbcabac+abc,=(a+bab)+cc(a+bab),=(a+bab)c,=(ab)c.

Therefore, (G,) is associative.

DoesG contain the identity e? Consider 0G.

a0=a+0a0,=a+00,=a.0a=0+a0a,=0+a0,=a.

Sincea0=0a=a for any aG, and 0G;e=0 is the identity of  (G,)

Does each element of G have an inverse in G? Let's assume for contradiction that a1G exists for all aG. Then we can say that:

aa1=0,a+a1aa1=0,a1(1a)=a,a1=a1a.

Now consider where a=1. In this case, 11=111=10. Clearly 10G, so11G. This is a contradiction. Therefore, there exists aG such that a1G

Since(G,) is closed, has an identity, is associative, but does not contain an inverse for each element; (G,) is not a group.\ ((G, \star)\) is a monoid.

2. Let G:=3Z={3m:mZ}. Let m,n,pZ. Clearly 3m,3n,3p3Z by definition of 3Z. Consider (G,+)

Is (G,+) closed? Consider the following:

3m+3n=3(m+n).

Since Z is closed under addition, (m+n)Z, and 3(m+n)G. Therefore, (G,+) is closed. 

Is (G,+) associative? Consider the following:

3m+(3n+3p)=3(m+(n+p)),=3((m+n)+p),=(3m+3n)+3p.

Therefore, (G,+) is associative. 

DoesG contain the identity e? Consider 0G.

3m+0=3m.0+3m=3m.

Clearly 3m+0=0+3m=3m. Therefore, e=0 is the identity of (G,+).

Does(G,+) contain the inverse for each element inG? Consider 3(n) as the inverse for 3n. Since nZ, 3(n)G.

3n+3(n)=3(nn)=3(0)=0=e. and 3(n)+3n=3((n)+n)=3(0)=0=e.

Therefore 3n+3(n)=3(n)+3n=e. So for any aG, there exists a1G

Since (G,+) is closed, associative, and contains the identity and an inverse for every element, (G,+) is a group.

3. The Cayley table of G with multiplication mod(20):

(mod20) 4 8 12 16
4 16 12 8 4
8 12 4 16 8
12 8 16 4 12
16 4 8 12 16

Is (G,(mod(20)) closed? Yes, we can see from the Cayley table that all results of (mod(20) are in G

Does G contain the identity for (mod(20) ? Again, by the Cayley table, we can see that the identity e=16

Is (G,(mod(20)) associative? Yes. We know by the properties of Z that multiplication of integers is associative. Also, by the properties of modular arithmetic that the modulo operation has no impact on associativity. Therefore, (G,(mod(20)) is associative. \\

Does G contain an inverse for each element under (mod(20)? Yes, consider the following:

4416mod(20)41=4.81216mod(20)81=12.12816mod(20)121=8.161616mod(20)161=16.

So clearly for each aG, there exists a1G. Since (G,(mod(20)) is closed, associative, and contains the identity and an inverse for every element, (G,(mod(20)) is a group.

 

Exercise 2

Decide whether the following structures form a group or not by looking at the Cayley table.

  1.  {±1,±i} with  multiplication, Where i is the complex number such that i2=1.
  2. 1=[1001], I=[0110], J=[0ii0], K=[i00i], where i is the complex number such that i2=1.

Define  Q8={±1,±I,±J,±K}.  Q8  with matrix multiplication.
 

Answer

1. Let G={1,1, i,i} with complex multiplication.

We are to determine whether the set G forms a group (G,) under multiplication, where i is the imaginary unit with i2=1. We will construct and examine the Cayley table to verify the group properties.

clipboard_e5404ae73bf4921de2ebadc615e60c2ab.png 

Since it is closed, has an identity, and every element has an inverse according to the table, multiplication is associative on 2×2 matrices over the complex numbers; it is a group.

2.  Q8={±1,±I,±J,±K}, where 1=[1001], I=[0110], J=[0ii0], K=[i00i], where i is the complex number such that i2=1.

clipboard_e774f496b1bece3db03807b68dd302e70.png

Since it is closed, has an identity, and every element has an inverse according to the table, multiplication is associative on 2×2 matrices over the complex numbers; it is a group.

 

 

Exercise 3

Prove or disprove:  Let H and K be subgroups of the group G. Then

  1. HK is a subgroup of G.
  2.  HK is a subgroup of G.
Answer

1. No, HK is a subgroup of G.clipboard_eac8a75090628659f9ced271e53de278e.png

  Answers may vary.

Counterexample:

Let G=(Z6,+(mod6))

Thus eG=0.

We shall show that HK is not a subgroup of G.

Consider Z6={0,1,2,3,4,5}.

Let H=({0,2,4},+(mod6)).   

Note HG  sinceeG=0H and 1h1,hH.

Let K=({0,3},+(mod6)).   

Note KG since eK=0H and 1k1,kK.

Thus HK=({0,2,3,4},+(mod6)).

Let a=2 andb=3, where 2,3HK.

However, since 2+35(mod6)HK, HK is not a group.

Thus, HK is not a subgroup ofG.

2.  Yes, HK is a subgroup of G .

Proof:

Let e be the identity of G .clipboard_e9ee44946ec695b521b9db9c118458312.png

Then e is the identity of both H and K .

Therefore eHK .

Let g1,g2HK .

Then we shall show that g1g12HK .

Since g1,g2HK , g1,g2H and g1,g2K .

Since HG and KG , g1,g2,g11,g12 are in both H and K .

Hence g1g12HK .

Therefore HKG .◻

Exercise 4

 1. Show that U(20)≠<k> for any k in U(20). [Hence U(20) is not cyclic.]

2. Prove that if  G is an abelian group, then for each element aGC(a)={xG|xa=ax} is a subgroup of G.

Answer

1. 

U(20)={1,3,7,9,11,13,17,19}  and |U(20)|=8 .

We shall show thatU(20)k,kU(20) .

 

k

Calculations

Order

 ofk

1

11(mod20)

1

3

313(mod20) ;329(mod20) ;337(mod20) ;341(mod20)

4

7

717(mod20) ;729(mod20) ;733(mod20) ;741(mod20)

4

9

919(mod20) ;921(mod20)

2

11

11111(mod20) ;1121(mod20)

2

13

13113(mod20) ;1329(mod20) ;13317(mod20) ;1341(mod20)

4

17

17117(mod20) ;1729(mod20) ;17313(mod20) ;1741(mod20)

4

19

19119(mod20) ;1921(mod20)

2

 

Since |k||U(20)|,kU(20) , there is no kU(20) that generates U(20) .

2. Since G is abelian,  for each aGC(a)={xG|xa=ax}=G is a subgroup of G.

 

 


This page titled Assignment 3 is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Pamini Thangarajah.

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